NEET Physics Class 12 Notes For Chapter 1 Magnetic Field

Magnetic Effect Of Current And Magnetic Force On Charge Or Current

Magnet:

Even after being neutral (showing no electric interaction), two bodies may attract/repel strongly if they have a special property. This property is known as magnetism. This force is called magnetic force.

Those bodies are called magnets. Later on, we will see that it is due to circulating currents inside the atoms. Magnets are found in different shapes, but a bar magnet is frequently used for many experimental purposes.

When a bar magnet is suspended at its middle, as shown, and it is free to rotate in the horizontal plane it always comes to equilibrium in a fixed direction.

One end of the magnet (say A) is directed approximately towards the north and the other (say B) approximately towards the south. This observation is made everywhere on the earth.

Due to this reason the end A, which points towards the north direction is called NORTH POLE and the other end which points towards the south direction is called the SOUTH POLE. They can be marked as ‘N’ and ‘S’ on the magnet.

This property can be used to determine the north or south direction anywhere on the earth and indirectly east and west also if they are not known by another method (like rising of sun and setting of the sun).

This method is used by navigators of ships and aeroplanes. The directions are as shown in the figure. In all directions, E, W, N, and S are in the horizontal plane.

The magnet rotates due to the earth’s magnetic field which we will discuss later in this chapter.

Pole Strength Magnetic Dipole And Magnetic Dipole Moment

A magnet always has two poles ‘N’ and ‘S’ and like poles of two magnets repel each other and the unlike poles of two magnets attract each other they form an action-reaction pair.

The poles of the same magnet do not come to meet each other due to attraction. They are maintained we cannot get two isolated poles by cutting the magnet from the middle. The other end becomes a pole of the opposite nature.

So, ‘N’ and ‘S’ always exist together therefore they are Known as +ve and –ve poles. The north pole is treated as a positive pole (or positive magnetic charge) and the south pole is treated as a –ve pole (or –ve magnetic charge).

They are quantitatively represented by their ”POLE STRENGTH” +m and –m respectively (just like we have charges +q and –q in electrostatics). Pole strength is a scalar quantity and represents the strength of the pole hence, of the magnet also).

A magnet can be treated as a dipole since it always has two opposite poles (just like in an electric dipole we have two opposite charges –q and +q). It is called a Magnetic Dipole and it has a Magnetic M Dipole Moment. It is represented by. It is a vector quantity. Its direction is from –m to +m which means from ‘S’ to ‘N’)

M = m.lm m here lm m = magnetic length of the magnet. Lm is slightly less than lg (it is the geometrical length of the magnet = end-to-end distance). The ‘N’ and ‘S’ are not located exactly at the ends of the magnet. For calculation purposes, we can assume lm = lg [Actually lm/lg ~ 0.84].

The units of m and M will be mentioned afterward so that you can remember and understand them.

Magnetic Field And Strength Of Magnetic Field.

The physical space around a magnetic pole has a special influence due to which other poles experience a force. That special influence is called magnetic field and that force is called ‘magnetic force’.

This field is qualitatively represented by the ‘strength of magnetic field’ or b “magnetic induction” or “magnetic flux density”. It is represented by \(\overrightarrow{\mathrm{b}}\). It is a vector quantity.

Definition of \(\overrightarrow{\mathrm{B}}\): The magnetic force experienced by a north pole of unit pole strength at a point due to some other poles (called source) is called the strength of the magnetic field at that point due to the source.

Mathematically \(\vec{B}=\frac{\vec{F}}{m}\)

Here \(\overrightarrow{\mathrm{F}}\) = magnetic force on the pole of pole strength m. m may be +ve or –ve and of any value. B S.\(\overrightarrow{\mathrm{B}}\). unit of is Tesla or Weber/m2 (abbreviated as T and Wb/m2).

We can also write \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{B}}\) According to this direction of on +ve pole (North pole) will be in the direction B of field and on –ve pole (south pole) it will be opposite to the direction of \(\overrightarrow{\mathrm{B}}\)

The field generated by sources does not depend on the test pole (for its value and any sign).

⇒ \(\overrightarrow{\mathrm{B}}\) due to various source

Due to a single pole:

(Similar to the case of a point charge in electrostatics)

⇒ \(B=\left(\frac{\mu_0}{4 \pi}\right) \frac{m}{r^2} \text {. }\)

This is magnitude

The direction of B due to the north pole and south poles are as shown

in vector form \(\vec{B}=\left(\frac{\mu_0}{4 \pi}\right) \frac{m}{r^3} \vec{r}\) here m is with a sign and = position vector of the test point for the pole.

Due to a bar magnet:

(Same as the case of electric dipole in electrostatics) Independent case never found. Always ‘N’ and ‘S’ exist together as magnets.

at A (on the axis) \(=\left(\frac{\mu_0}{4 \pi}\right) \frac{\vec{M}}{r^3} \quad \text { for } \quad a \ll r\)

at B (on the equatorial) \(=-\left(\frac{\mu_0}{4 \pi}\right) \frac{\vec{M}}{r^3} \text { for } \mathrm{a} \ll \mathrm{r}\)

At General Point

⇒ \(B_r=2\left(\frac{\mu_0}{4 \pi}\right) \frac{M \cos \theta}{r^3} \quad \Rightarrow \quad B_n=2\left(\frac{\mu_0}{4 \pi}\right) \frac{M \sin \theta}{r^3}\)

⇒ \(\begin{aligned}
& B_{\text {res }}=\frac{\mu_0 M}{4 \pi r^3} \sqrt{1+3 \cos ^2 \theta} \\
& \tan \phi=\frac{B_n}{B_r}=\frac{\tan \theta}{2}
\end{aligned}\)

Magnetic lines of force of a bar magnet:

Solved Examples

Example 1. Find the magnetic field due to a dipole of magnetic moment 1.2 A-m2 at a point 1 m away from it in a direction making an angle of 60º with the dipole axis.
Solution: The magnitude of the field is

⇒ \(B=\frac{\mu_0}{4 \pi} \frac{M}{r^3} \sqrt{1+3 \cos ^2 \theta}\)

⇒ \(=\left(10^{-7} \frac{\mathrm{T}-\mathrm{m}}{\mathrm{A}}\right) \frac{1.2 \mathrm{~A}-\mathrm{m}^2}{1 \mathrm{~m}^3} \sqrt{1+3 \cos ^2 60^{\circ}}=1.6 \times 10^{-7} \mathrm{~T} \text {. }\)

The direction of the field makes an angle with the radial line where

⇒ \(\tan \alpha=\frac{\tan \theta}{2}=\frac{\sqrt{3}}{2}\)

Example 2. The figure shows two identical magnetic dipoles a and b of magnetic moments M each, placed at a separation d, with their axes perpendicular to each other. Find the magnetic field at the point P midway between the dipoles.

Solution: Point P is in the end-on position for dipole a and in the broadside-on position for dipole b.

The magnetic field at p due to a is \(\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{(\mathrm{d} / 2)^3}\) along the axis of a, and that due to b is \(B_b=\frac{\mu_0}{4 \pi} \frac{M}{(d / 2)^3}\) parallel to the axis of b as shown in figure. The resultant field at P is, therefore.

⇒ \(B=\sqrt{B_a^2+B_b^2}=\frac{\mu_0 M}{4 \pi(d / 2)^3} \sqrt{1^2+2^2}\)

⇒ \(=\frac{2 \sqrt{5} \mu_0 \mathrm{M}}{\pi d^2}\)

The direction of this field makes an angle α with Ba such that tanθ = Bb/Ba = 1/2.

Magnet in an external uniform magnetic field

(same as the case of the electric dipole)

Here θ is angle between and \(\vec{B} \text { and } \vec{M}\)

Note:

⇒ \(\vec{\tau}\) acts such that it tries to make \(\overrightarrow{\mathrm{M}} \times \vec{B}\)

⇒ \(\vec{\tau}\) is the same about every point of the dipole’s potential energy is

U = – MB cos θ = \(\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}\)

θ = 0º is stable equilibrium

θ = π is an unstable equilibrium

For small ‘θ’ the dipole performs SHM about θ = 0º position

t= 0– MB sin θ ;

I α = – MB sin θ

for small θ, \(\sin \theta \simeq \theta\) \(\alpha=-\left(\frac{\mathrm{MB}}{\mathrm{I}}\right) \theta\)

Angular frequency of SHM

⇒ \(\omega=\sqrt{\frac{M B}{I}}=\frac{2 \pi}{T} \quad \Rightarrow \quad T=2 \pi \sqrt{\frac{I}{M B}}\)

Here = \(I_{c m}\) if the dipole is free to rotate

=\(I_{hinge}\) if the dipole is hinged.

Example 3. A bar magnet having a magnetic moment of 1.0 × 10^-4 J/T is free to rotate in a horizontal plane. A horizontal magnetic field B = 4 × 10^-5 T exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction 60º from the field.
Solution: The work done by the external agent = change in potential energy
= (–MB cosθ₂) –(–MB cosθ₁) = –MB (cos 60º – cos 0º)

⇒ \(=\frac{1}{2} M B=\frac{1}{2} \times\left(1.0 \times 10^4 \mathrm{~J} / \mathrm{T}\right)\left(4 \times 10^{-5} \mathrm{~T}\right)=0.2 \mathrm{~J}\)

Example 4. A magnet of magnetic dipole moment M is released in a uniform magnetic field of induction B from the position shown in the figure. Find :

1. Its kinetic energy at θ = 90º
2. Its maximum kinetic energy during the motion.
3. Will it perform SHM? Oscillation? Periodic motion? What is its amplitude?

Solution: Apply energy conservation at θ= 120º and θ= 90º

= – MB cos 90º + (K.E.)

⇒ \(\mathrm{KE}=\frac{\mathrm{MB}}{2}\)

⇒ \(\mathrm{KE}=\frac{\mathrm{MB}}{2}\)

K.E. will be the maximum whereas P.E. is the minimum. P.E. is minimum at θ= 0º. Now apply energy conservation between θ= 120º and θ= 0º. – MB cos 120º + 0 = –mB cos 0º + (KE)_max

⇒ \((\mathrm{KE})_{\max }=\frac{3}{2} \mathrm{MB}\)

The K.E. is max at θ= 0º can also be proved by the torque method. From θ= 120º to θ= 0º the torque always acts on the dipole in the same direction (here it is clockwise) so its K.E. keeps on increasing till θ= 0º. Beyond that θ reverses its direction and then K.E. starts decreasing

θ = 0º is the orientation of M to here the maximum K.E.

Since ‘θ’ is not small.

θ = 0º is the orientation of M to here the maximum K.E.

Since ‘θ’ is not small.

Magnet In An External Nonuniform Magnetic Field

No special formulas are applied is such problems. Instead, see the force on individual poles and calculate the resistant force torque on the dipole.

Magnetic effects of current (and moving charge)

It was observed by OERSTED that a current-carrying wire produces a magnetic field near it.

It can be tested by placing a magnet in the nearby space, it will show some movement (deflection or rotation of displacement). This observation shows that a current or moving charge produces a magnetic field.

Oersted Experiment And Observations

Oersted experimented in 1819 whose arrangement is shown in the following figure. The following observations were noted from this experiment.

1. When no current is passed through the wire AB, the magnetic needle remains undeflected.
2. When current is passed through the wire AB, the magnetic needle gets deflected in a particular direction and the deflection increases as the current increases.
3. When the current flowing in the wire is reversed, the magnitude needle gets deflected in the opposite direction and its deflection increases as the current increases.
   1. Oersted concluded from this experiment that on passing a current through the conducting wire, a magnetic field is produced around this wire. As a result, the magnetic needle is deflected. This phenomenon is called the magnetic effect of current.
   2. In another experiment, it was found that the magnetic lines of force due to the current flowing in the wire are in the form of concentric circles around the conducting wire.

Frame Dependence Of \(\overrightarrow{\mathrm{B}}\)

The motion of anything is a relative term. A charge may appear at rest by an observer (say O₁) and \(\overrightarrow{\mathrm{v}}_1\)  1 moving at the same velocity concerning observer O₂ and at velocity concerning observer O₃ then due to that charge w.r.t. O₁ will be zero and w.r. to O₂ and O₃ it will be and (that means different).

In a current-carrying wire electrons move in the opposite direction to that of the current and +ve ions (of the metal) are static w.r.t. the wire.

Now if some observer (O₁) moves with velocity Vd in the direction of motion of the electrons then electrons will have zero velocity and +ve ions will have velocity Vd in the downward direction w.r.t. O₁. The density (n) of +ve ions is the same as the density of free electrons and their charges are of the same magnitudes.

So, w.r.t. O₁ electrons will produce zero magnetic field but +ve ions will produce +ve same due to the current carrying wire does not depend on the reference frame (this is true for any velocity of the observer).

\(\overrightarrow{\mathrm{B}}\) due to magnet:

B produced by the magnet does not contain the term of velocity
B So, we can say that the due magnet does not depend on the frame.

\(\overrightarrow{\mathrm{B}}\) Due To A Point Charge

r A charge particle ‘q’ has velocity v as shown in the figure. It is at position ‘A’ at some time. Is the position vector of point ‘P’ w.r.

To position of the charge. Then at P due to q is

⇒ \(B=\left(\frac{\mu_0}{4 \pi}\right) \frac{q v \sin \theta}{r^2}\); here θ angle between \(\vec{v} \text { and } \vec{r}\)

⇒ \(\vec{B}=\left(\frac{\mu_0}{4 \pi}\right) \frac{q \vec{V} \times \vec{r}}{r^3}\) ; q with sign \(\vec{B} \perp \vec{v} \text { and also } \vec{B} \perp \vec{r} \text {. }\).

The direction will be found by using the rules of vector product.

Self Practice Problems

Question 1. The magnetic field is produced by the flow of current in a straight wire. This phenomenon is based on-

1. Faraday’s Law
2. Maxwell’s Law
3. Coulbom’s Law
4. Oersted’s Law

Answer: 4. Oersted’s Law

Question 2. The field produced by a moving charged particle is-

1. Electric
2. Magnetic
3. Both electric and magnetic
4. Nothing can be predicted

Answer: 3. Both electric and magnetic

Question 3. The magnetic field due to a small bar magnet at a distance varies as

1. \(\frac{1}{d^2}\)
2. \(\frac{1}{d^{3 / 2}}\)
3. \(\frac{1}{d^{3 / 2}}\)
4. \(\frac{1}{d}\)

Answer: 3. \(\frac{1}{d}\)

Question 4. A magnetic dipole of magnetic moment M is situated with its axis along the direction of a magnetic field of strength B. How much work will have to be done to rotate it through 180°?

1. -MB
2. +MB
3. Zero
4. +2MB

Answer: 4. +2MB

Question 5. The magnetic field due to a small magnetic dipole of magnetic moment M, at distance r from the centre on the equatorial line, is given by- (in the MKS system)

1. \(\frac{\mu_0}{4 \pi} \times \frac{M}{r^2}\)
2. \(\frac{\mu_0}{4 \pi} \times \frac{M}{r^3}\)
3. \(\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^2}\)
4. \(\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3}\)

Answer: 2. \(\frac{\mu_0}{4 \pi} \times \frac{M}{r^3}\)

Biot-Savant’s Law ( Due To A Wire)

It is an experimental law. A current ‘i’ flows in a wire (may be straight or curved). Due to θ the length of the wire the magnetic field at ‘P’ is

⇒ \(\mathrm{dB} \propto \text { id } \ell \quad \Rightarrow \quad \propto \frac{1}{\mathrm{r}^2} \quad \Rightarrow \quad \propto \sin \theta \quad \Rightarrow \quad \mathrm{dB} \propto \frac{\text { id } \ell \sin \theta}{\mathrm{r}^2}\)

⇒ \(\mathrm{dB}=\left(\frac{\mu_0}{4 \pi}\right) \frac{\mathrm{id} \ell \sin \theta}{\mathrm{r}^2} \quad \Rightarrow \quad \overrightarrow{\mathrm{dB}}=\left(\frac{\mu_0}{4 \pi}\right) \frac{\overrightarrow{i d} \times \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}\)

Here = position vector of the test point w.r.t. \(\overrightarrow{\mathrm{d} \ell}\)

α = angle between \(\overrightarrow{\mathrm{d} \ell} \text { and } \overrightarrow{\mathrm{r}}\). The Resultant \(\overrightarrow{\mathrm{B}}=\int \overrightarrow{\mathrm{dB}}\)

Using this fundamental formula we can derive the expression of \(\vec{B}\) due to a long wire.

⇒ \(\vec{B}\) due to a straight wire:

Due to a straight wire ‘PQ’ carrying a current ‘i’ the \(\vec{B}\) at A is given by the formula \(B=\frac{\mu_0 I}{4 \pi \mathrm{r}}\left(\sin \theta_1+\sin \theta_2\right)\)

(Derivation can be seen in a standard textbook like your school book or concept of physics of HCV part 2) Direction:

B Due to every element of ‘PQ’ at A being directed inwards. So its resultant is also directed inwards. It is represented by (x)

The direction at various points is shown in the figure shown.

At points ‘C’ and ‘D’ = 0 (think how). For the case shown in figure B at \(A=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_2-\sin \theta_1\right)\)

Shortcut for Direction:

The direction of the magnetic field at point P due to a straight wire can be found by a slight variation in the right-hand thumb rule. If we stretch the thumb of the right hand along the current and curl our fingers to pass through point P, the direction of the fingers at P gives the direction of the magnetic field there.

We can draw magnetic field lines on the pattern of electric field lines. A tangent to a magnetic field line given the direction of the magnetic field existing at that point.

For a straight wire, the field lines are concentric circles with their centers on the wire and in the plane perpendicular to the wire. There will be an infinite number of such lines in the planes parallel to the above-mentioned plane.

Example 5. Find the resultant magnetic field at ‘C’ in the figure shown.

Solution: It is clear that ‘B’ is at ‘C’ due to all the wires being directed. Also, B at ‘C’ due to PQ and SR are the same. Also due to QR and PS being the same

⇒ \(B_{r e s}=2\left(B_{P Q}+B_{S P}\right) \Rightarrow \quad B_{P Q}=\frac{\mu_0 \mathrm{i}}{4 \pi \frac{a}{2}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right),\)

⇒ \(B_{s p}=\frac{\mu_0 i}{4 \pi \frac{\sqrt{3} a}{2}}\left(\sin 30^{\circ}+\sin 30^{\circ}\right) \Rightarrow B_{r e s}=2\left(\frac{\sqrt{3} \mu_0 i}{2 \pi a}+\frac{\mu_0 i}{2 \pi a \sqrt{3}}\right)=\frac{4 \mu_0 i}{\sqrt{3} \pi a}\)

Example 6. The figure shows a square loop made from a uniform wire. Find the magnetic field at the centre of the square if a battery is connected between points A and C.

Solution: The current will be equally divided at A. The fields at the center due to the currents in the wires AB and DC will be equal in magnitude and opposite in direction. The result of these two fields will be zero. Similarly, the resultant of the fields due to the wires AD and BC will be zero. Hence, the net field at the center will be zero.

Special case:

If the wire is infinitely long then the magnetic field at ‘P’ (as shown in the figure) is given by (using θ1 = θ2 = 90º and the formula of ‘B’ due to straight wire.

⇒ \(B=\frac{\mu_0 I}{2 \pi r} \quad \Rightarrow \quad B \propto \frac{I}{r}\)

The direction of at various is as shown in the figure. The magnetic lines of force will be concentric circles around the wire (as shown earlier)

If the wire is infinitely long but ‘P’ is as shown in the figure. The B direction at various points is as shown in the figure. At ‘P’.

⇒ \(B=\frac{\mu_0 I}{4 \pi r}\)

Self Practice problems 

Question 6. A pair of stationary and infinitely long bent wires are placed in the xy plane as shown in The wires carry a current of i = 10 Amp. each as shown. The segments L and M are along the x-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m. The magnetic induction at the origin O is

1. \(10^{-4} \mathrm{~Wb} / \mathrm{m}^2\)
2. \(10^{-2} \mathrm{~Wb} / \mathrm{m}^2\)
3. \(10^{-6} \mathrm{~Wb} / \mathrm{m}^2\)
4. \(10^{-8} \mathrm{~Wb} / \mathrm{m}^2\)

Answer: 1. \(10^{-4} \mathrm{~Wb} / \mathrm{m}^2\)

Question 7. A current-carrying wire is bent into the shape of a square coil. The magnetic field produced at the center of the coil by one arm BC is B. Then the resultant magnetic field at the center due to all the arms will be

1. 4B
2. \(\frac{B}{2}\)
3. B
4. \(\frac{2}{B}\)

Answer: 1. 4B

Question 8. As shown in the diagram, two perpendicular wires are placed very close to each other, but they are not touching each other. The points where the intensity of the magnetic field is zero are-

1. A
2. B, D
3. A, B
4. B

Answer: 2. B, D

Question 9. At a distance of 10 cm. from a long straight wire carrying current, the magnetic field is 0.04. Tesla the magnetic field at the distance of 40 cm. will be

1. 0.001T
2. 0.02T
3. 0.08T
4. 0.16T

Answer: 1. 0.001T

⇒ \(\stackrel{\rightharpoonup}{B}\) due to circular loop

At centre: Due to each \(\overrightarrow{\mathrm{d} \ell}\) element of the loop \(\stackrel{\rightharpoonup}{B}\) At ‘c’ is inwards (in this case)

⇒ \(\overline{B_{r e s}} \text { at ‘ } c \text { ‘ is } \otimes . B=\frac{\mu_0 N I}{2 R} \text {, }\)

N = No. of turns in the loop =\(\frac{\ell}{2 \pi R}\) l= length of the loop.

N can be fraction \(\left(\frac{1}{4}, \frac{1}{3}, \frac{11}{3} \text { etc. }\right)\) or integer.

The direction of \(\stackrel{\rightharpoonup}{B}\): The direction of the magnetic field at the center of a circular wire can be obtained using the right-hand thumb rule. If the fingers are curled along the current, the stretched thumb will point toward the magnetic field.

Another way to find the direction is to look into the loop along its axis. If the current is in an anticlockwise direction, the magnetic field is towards the viewer. If the current is in a clockwise direction, the field is away from the viewer.

Semicircular And Quarter Of A Circle:

On the axis of the loop= \(B=\frac{\mu_0 N^2 R^2}{2\left(R^2+x^2\right)^{3 / 2}}\)

Direction can be obtained by the right-hand and thumb rule. Curl your fingers in the direction of the current then B the direction of the thumb points in the direction of at the points on the axis.

The magnetic field at a point not on the axis is mathematically difficult to calculate. We show qualitatively in the figure the magnetic field lines due to a circular current which will give some idea of the field.

A loop as a magnet:

The pattern of the magnetic field is comparable with the magnetic field produced by a bar magnet.

The side ‘I’ (the side from which the emerges out) of the loop acts as the ‘NORTH POLE’ and side II (the B side in which the enters) acts as the ‘SOUTH POLE’. It can be verified by studying the force on one loop due to a magnet or a loop.

Mathematically:

⇒ \(B_{\text {amis }}=\frac{\mu_0 N I R^2}{2\left(R^2+x^2\right)^{3 / 2}} \cong \frac{\mu_0 N I R^2}{2 x^3} \text { for } x>R=2\left(\frac{\mu_0}{4 \pi}\right)\left(\frac{I N \pi R^2}{x^3}\right)\)

It is similar to Baxis due to magnet \(=2\left(\frac{\mu_0}{4 \pi}\right) \frac{m}{x^3}\)

Magnetic dipole moment of the loop

M = INπR²

M = INA for any other shaped loop.

Unit of M is Amp. m²

Unit of m (pole strength) = Amp. m {since in magnet M = ml}

⇒ \({\mathrm{M}}=\mathrm{IN} \overrightarrow{\mathrm{A}} \text {, }\)

⇒ \(\vec{A}\) = unit normal vector for the loop.
To be determined by right hand rule which is also used to determine the direction of the axis. It is also from

‘S’ side to ‘N’ side of the loop.

Solenoid:

A solenoid contains a large number of circular loops wrapped around a non-conducting cylinder. (it may be a hollow cylinder or it may be a solid cylinder)

The winding of the wire is the uniform direction of the magnetic field is the same at all points of the axis.

⇒ \(\vec{B}\) on axis (turns should be very close to each other).

⇒ \(B=\frac{\mu_0 n i}{2}\left(\cos \theta_1-\cos \theta_2\right)\)

where n: number of turns per unit length.

⇒ \(\cos \theta_1=\frac{\ell_1}{\sqrt{\ell_1^2+\mathrm{R}^2}} ; \quad \cos \beta=\frac{\ell_2}{\sqrt{\ell_2^2+\mathrm{R}^2}}=-\cos \theta_2\)

⇒ \(B=\frac{\mu_0 n i}{2}\left[\frac{\ell_1}{\sqrt{\ell_1^2+R^2}}+\frac{\ell_2}{\sqrt{\ell_2^2+R^2}}\right]=\frac{\mu_0 n i}{2}\left(\cos \theta_1+\cos \beta\right)\)

Note: Use the right-hand rule for direction (same as the direction due to the loop).

Derivation:

Take an element of width dx at a distance x from point P. [point P is the point on the axis at which we are going to calculate the magnetic field. Total number of turns in the element dn = ndx where n: number of turns per unit length.

⇒ \(d B=\frac{\mu_0 i R^2}{2\left(R^2+x^2\right)^{3 / 2}}(n d x)\)

⇒ \(B=\int d B=\int_{-\ell_1}^2 \frac{\mu_0 i R^2 n d x}{2\left(R^2+x^2\right)^{3 / 2}}=\frac{\mu_0 n i}{2}\left[\frac{\ell_1}{\sqrt{\ell_1^2+R^2}}+\frac{\ell_2}{\sqrt{\ell_2^2+R^2}}\right]=\frac{\mu_0 n i}{2}\left[\cos \theta_1-\cos \theta_2\right]\)

For ‘Ideal Solenoid’:

Inside (at the midpoint)

⇒ \(\begin{aligned}
& \ell>R \quad \text { or length is infinite } \\
& \theta_1 \rightarrow 0 \\
& \theta_2 \rightarrow \pi \\
& B=\frac{\mu_0 n i}{2}[1-(-1)] \\
& \mathbf{B}=\mu_0 \mathbf{n i}
\end{aligned}\)

If the material of the solid cylinder has relative permeability \(\text { ‘ } \mu_t^{\prime} \text { then } B=\mu_0 \mu_t \text { ni }\)

At the ends B \(B=\frac{\mu_0 n i}{2}\)

Comparison between ideal and real solenoid:

1. Ideal Solenoid Real Solenoid

Self Practice Problems

Question 10. The radius of a circular coil is R and it carries a current of I ampere. The intensity of the magnetic field at a distance x from the centre (x >> R) will be:

1. \(\mathrm{B}=\frac{\mu_0}{2} \frac{\mathrm{IR}^2}{\mathrm{x}^2}\)
2. \(B=\frac{\mu_0 I^2}{2 x^3}\)
3. \(B=\frac{\mu_0 I R}{2 x^2}\)
4. \(B=\frac{\mu_0 \text { IR }}{2 x^3}\)

Answer: 2. \(B=\frac{\mu_0 I^2}{2 x^3}\)

Question 11. The magnetic field B at the centre of a circulate coil of radius r is times that due to a long straight wire at a distance r from it, for equal currents. Fig shows three cases; in all cases, the circular part has a radius of r and the straight ones are infinitely long. For the same current the field B at the centre P in cases 1,2, and 3 has the ratio.

Answer: 2. \(B=\frac{\mu_0 I^2}{2 x^3}\)

Question 12. If the intensity of the magnetic field at a point on the axis of the current-carrying coil is half of that at the centre of the coil, then the distance of that point from the centre of the coil will be

1. \(\left(-\frac{\pi}{2}\right): \frac{\pi}{2}:\left[\frac{3 \pi}{4}-\frac{1}{2}\right]\)
2. \(\left(-\frac{\pi}{2}+1\right):\left[\frac{\pi}{2}+1\right]:\left[\frac{3 \pi}{4}+\frac{1}{2}\right]\)
3. \(-\frac{\pi}{2}: \frac{\pi}{2}: \frac{3 \pi}{4}\)
4. \(\left(-\frac{\pi}{2}-1\right):\left[\frac{\pi}{4}-\frac{1}{4}\right]:\left[\frac{3 \pi}{4}+\frac{1}{2}\right]\)

Answer: 1. \(\left(-\frac{\pi}{2}\right): \frac{\pi}{2}:\left[\frac{3 \pi}{4}-\frac{1}{2}\right]\)

Question 13. If the intensity of the magnetic field at a point on the axis of current carrying coil is half of that at the centre of the coil, then the distance of that point from the centre of the coil will be

1. \(\frac{R}{2}\)
2. R
3. \(\frac{3 R}{2}\)
4. 0.766R

Answer: 4. \(\frac{3 R}{2}\)

Question 13. A current of 0.1 ampere flows through a coil of 100 turns and the radius is 5 cm. The magnetic field at the centre of the coil will be

1. \(4 \pi \times 10^{-5} \mathrm{~T}\)
2. \(8 \pi \times 10^{-5} \mathrm{~T}\)
3. \(4 \times 10^{-5} \mathrm{~T}\)
4. \(2 \times 10^{-5} \mathrm{~T}\)

Answer: 1. \(4 \pi \times 10^{-5} \mathrm{~T}\)

Question 14. A current I flows through a circular coil of radius r the intensity of the field at its centre is

1. Proportional to r
2. Inversely proportional I
3. Proportional to I
4. Proportional to I2

Answer: 3. Proportional to I

Question 15. In a current carrying long solenoid the field produced does not depend upon-

1. Number of turns per unit length
2. Current flowing
3. The radius of the solenoid
4. All of the above three

Answer: 3. Radius of the solenoid

Ampere’s Circuital Law :

The line integral \(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}\) on a closed curve of any shape is equal to (permeability of free space) times the net current through the area bounded by the curve.

⇒ \(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\mu_0 \Sigma \mathrm{I}\)

Note:

A line integral is independent of the shape of the path and the position of the within in it.

The statement \(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=0\) does not necessarily mean that \(\vec{B}=0\) everywhere along the path but only that no net current is passing through the path.

Sign of current: The current due to which \(\vec{B}\) is produced in the same sense as \(\overrightarrow{\mathrm{d} \ell}\) (i.e \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}\) positive will be taken positive and the current which produces \(\vec{B}\) in the sense opposite to \(\overrightarrow{\mathrm{d} \ell}\) will be negative

Solved Examples

Example 7. Find the values of \(\int \vec{B} \cdot \overrightarrow{d \ell}\) for the loops L1, L2, and L3 in the figure shown. The sense of \(\overrightarrow{\mathrm{d} \ell}\) is mentioned in the figure.

Solution: \(\text { for } L_1 \int \vec{B} \cdot \overrightarrow{d \ell}=\mu_0\left(I_1-\mathrm{I}_2\right)\)

here I1 is taken positively because magnetic lines of force produced by I1 are anti-clockwise as B d seen from the top. I2 produces lines of \(\overrightarrow{\mathrm{B}}\) in a clockwise sense as seen from the top. The sense of is anticlockwise as seen from the top.

For \(L_2: \int \vec{B} \cdot \vec{d}=\mu_0 \quad\left(I_1-I_2+I_4\right)\) for \(\mathrm{L}_3: \int \vec{B} \cdot \vec{d} \ell=0\)

To find out the magnetic field due to infinite current carrying wire

By B.S.L \(\overrightarrow{\mathrm{B}}\) will have circular lines. \(\overrightarrow{d \ell}\) is also taken tangent to the circle.

⇒ \(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\) since \(\theta=0^{\circ} \text { so } \mathrm{B} \int \mathrm{d} \ell=\mathrm{B} 2 \pi \mathrm{R}\) (since B= const)

Now by amperes law: \(\text { B } 2 \pi R=\mu_0 I\) since \(B=\frac{\mu_0 i}{2 \pi R}\)

Hollow current carrying infinitely long cylinder : (I is uniformly distributed on the whole circumference)

For r > R By symmetry, the amperian loop is a circle.

⇒ \(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\int \mathrm{B} \mathrm{d} \ell\) since 0=0

⇒ \(=\mathrm{B} \int_0^2 \mathrm{~d} \ell\) since b= const. \(B=\frac{\mu_0 I}{2 \pi r}\)

r<R = \(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}=\int \mathrm{B} d \ell=\mathrm{B}(2 \pi \mathrm{r})=0\) = \(B_{\text {in }}=0\)

Graph:

Solid infinite current carrying cylinder:

Assume current is uniformly distributed on the whole cross-section area.

Current density \(\mathrm{J}=\frac{\mathrm{I}}{\pi \mathrm{R}^2}\)

Case \(r \leq R\)

Take an American loop inside the cylinder. By symmetry it should be a circle whose centre is on the axis of the cylinder and its axis also coincides with the cylinder axis on the loop.

⇒ \(\iint \vec{B} \cdot \vec{d} \ell=\int \mathrm{B} \cdot \mathrm{d} \ell=\mathrm{B} \int \mathrm{d} \ell=\mathrm{B} \cdot 2 \pi \mathrm{r}=\mu_0 \frac{1}{\pi \mathrm{R}^2} \pi \mathrm{r}^2\)

⇒ \(B=\frac{\mu_0 I r}{2 \pi R^2}=\frac{\mu_0 J r}{2} \quad \Rightarrow \quad \vec{B}=\frac{\mu_0(\vec{J} \times \vec{r})}{2}\)

Case 2: \(\int \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\int \mathrm{B} \mathrm{d} \ell=\mathrm{B} \int \mathrm{d} \ell=\mathrm{B} \cdot(2 \pi \mathrm{r})=\mu_0 \cdot \mathrm{I}\)

⇒ \(\Rightarrow \quad B=\frac{\mu_0 I}{2 \pi r} \text { also } \frac{\mu_0 I}{2 \pi r}(\hat{J} \times \hat{r})=\frac{\mu_0 J \pi R^2}{2 \pi r}\)

⇒ \(\vec{B}=\frac{\mu_0 R^2}{2 r^2}(\vec{J} \times \overrightarrow{\mathrm{r}})\)

Self Practice Problems

Question 16. The intensity of the magnetic field at a point situated at a distance r close to a long straight current carrying r wire is B the intensity of the field at a distance \(\frac{r}{2}\) from the wire will be-

1. \(\frac{B}{2}\)
2. \(\frac{B}{4}\)
3. 2B
4. 4B

Answer: 3. 2B

Question 17. Two straight infinitely long and thin wires are separated 0.1 m apart and carry a current of 10 Amp. each in opposite directions. The magnetic field on point at a distance of 0.1 m from both wires is

1. \(2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\)
2. \(3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\)
3. \(4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\)
4. \(1 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\)

Answer: 1. \(3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\)

Question 18. One ampere current is passed through a 2m long straight wire. The magnetic field in the air at a point distance of 3m from one end of the wire on its axis will be

1. \(\frac{\mu_0}{2 \pi}\)
2. \(\frac{\mu_0}{4 \pi}\)
3. \(\frac{\mu_0}{8 \pi}\)
4. zero

Answer: 4. Zero

Solved Examples 

Example 8. The figure shows a cross-section of a large metal sheet carrying an electric current along its surface. The current in a strip of width dl is Kdl where K is a constant. Find the magnetic field at a point P at a distance x from the metal sheet.

Solution: Consider two strips A and C of the sheet situated symmetrically on the two sides of P (figure). The magnetic field at P due to strip A is B 0 perpendicular to AP and that due to strip, C is BC perpendicular to CP. The resultant of these two is parallel to the width AC of the sheet. The field due to the whole sheet will also be in this direction. Suppose this field has magnitude B.

The field on the opposite side of the sheet at the same distance will also be B but in the opposite direction. Applying Ampere’s law to the rectangle shown in the figure.

⇒ \(2 \mathrm{~B} \ell=\mu_0 \mathrm{~K} \ell \quad \text { or, } \quad \mathrm{B}=\frac{1}{2} \mu_0 \mathrm{~K} \text {. }\)

Note that it is independent of x.

Example 9. Three identical long solenoids P, Q and R are connected as shown in the figure. If the magnetic field at the centre of P is 2.0 T, what would be the field at the centre of Q? Assume that the field due to any solenoid is confined within the volume of that solenoid only.

Solution: As the solenoids are identical, the currents in Q and R will be the same and will be half the current in P. The magnetic field within a solenoid is given by B = μ0ni. Hence the field in Q will be equal to the field in R and will be half the field in P i.e., will be 1.0 T.

Magnetic Field in a Toroid

1. A toroid is like an endless cylindrical solenoid, i.e. if a long solenoid is bent round in the form of a closed ring, then it becomes a toroid.
2. Electrically insulated wire is wound uniformly over the toroid as shown in the figure.
3. The thickness of the toroid is kept small in comparison to its radius and the number of turns is kept very large.
4. When a current i is passed through the toroid, each turn of the toroid produces a magnetic field along the axis at its centre. Due to the uniform distribution of turns this magnetic field has the same magnitude at their centres. Thus the magnetic lines of force inside the toroid are circular.
5. The magnetic field inside a toroid at all points is the same but outside the toroid, it is zero.
6. If the total number of turns in a toroid is N and R is its radius, then the number of turns per unit length of the toroid will be \(\mathrm{n}=\frac{\mathrm{N}}{2 \pi \mathrm{R}}\)
7. The magnetic field due to the toroid is determined by Ampere’s law.
8. The magnetic field due to toroid is \(B_0=\mu_0 \text { ni } \quad \text { or } \quad B_0=\mu_0\left(\frac{N}{2 \pi R}\right) \mathrm{i}\)
9. If a substance of permeability μ is placed inside the toroid, then B = uni
10. If ur is the relative magnetic permeability of the substance, then B = uru0 ni

Self Practice Problems

Question 19. A toroid has n turn density, current i then the magnetic field is

1. u0 ni
2. \(\mu_0 \frac{i}{n}\)
3. Zero
4. \(\mu_0 n^2 i\)

Answer: 1. u0 ni

Question 20. The current on the windings on a Toroid is 2.0 A. There are 400 turns and the mean circumferential length is 40 cm. If the inside magnetic field is 1.0 T, the relative permeability is near to

1. 100
2. 200
3. 300
4. 400

Answer: 4. 400

Current And Magnetic Field Due To Circular Motion Of A Charge

According to the theory of atomic structure every atom is made of electrons, protons and neutrons. Protons and neutrons are in the nucleus of each atom and electrons are assumed to be moving in different orbits around the nucleus.

An electron and a proton present in the atom constitute an electric dipole at every moment but the direction of this dipole changes continuously and hence at any time the average dipole moment is zero.

As a result static electric field is not observed.

Moving charge produces a magnetic field and the average value of this field in the atom is not zero.

In an atom, an electron moves in a circular path around the nucleus. Due to this motion current appears to be flowing in the electronic orbit and the orbit behaves like a current-carrying coil. If e is the electron charge, R is the radius of the orbit and f is the frequency of motion of an electron in the orbit, then

1. Current in the orbit = charge × frequency = ef
2. If T is the period, then \(f=\frac{1}{T}\) therefore \(\mathrm{i}=\frac{\mathrm{e}}{\mathrm{T}}\)
3. The magnetic field at the nucleus (centre) \(B_0=\frac{\mu_0 \mathrm{i}}{2 R}=\frac{\mu_0 \mathrm{ef}}{2 R}=\frac{\mu_0 \mathrm{e}}{2 R T}\)
4. If the angular velocity of the electron is, then \(\omega=2 \pi f \text { and } f=\frac{\omega}{2 \pi}\)
   • ∴ \(\mathrm{i}=\mathrm{ef}=\frac{\mathrm{e} \omega}{2 \pi}\)
   • ∴ \(B_0=\frac{\mu_0 i}{2 R}=\frac{\mu_0 \mathrm{e} \omega}{4 \pi R}\)
5. If the linear velocity of the electron is v, then v Rw R(2f) or \(f=\left(\frac{v}{2 \pi R}\right)\) or \(f=\left(\frac{v}{2 \pi R}\right)\)
   • ∴ \(i=e f=\frac{e v}{2 \pi R}\)
   • ∴ \(B_0=\frac{\mu_0 \mathrm{i}}{2 R}=\frac{\mu_0 e v}{4 \pi R^2}\)
6. Magnetic moment due to the motion of an electron in an orbit
   • \(M=i A=e f \pi R^2=\frac{e \pi R^2}{T}\) or, \(M=\frac{e \omega \pi R^2}{2 \pi}=\frac{e \omega R^2}{2}\) or, \(M=\frac{e v \pi R^2}{2 \pi R}=\frac{e v R}{2}\)

If the angular momentum of the electron is L, then
L mvR mR= mwR2

Writing M in terms of L

⇒ \(M=\frac{e m \omega R^2}{2 m}=\frac{e m v R}{2 m}=\frac{e L}{2 m}\)

According to Bohr’s second postulate \(m v R=n \frac{h}{2 \pi}\)

In ground state n = 1

⇒ \(L=\frac{h}{2 \pi}\)

∴ \(M=\frac{e h}{4 \pi m}\)

1. If a charge q (or a charged ring of charge q) is moving in a circular path of radius R with a frequency for angular velocity w, then
2. Current due to moving charge \(\mathrm{i}=\mathrm{qf}=\mathrm{q} \omega / 2 \pi\)
3. The magnetic field at the centre of the ring \(M=i\left(\pi R^2\right)=q f \pi R^2=\frac{1}{2} q \omega R^2\)
4. If a charge q is distributed uniformly over the surface of the plastic disc of radius R and it is rotated about its axis with an angular velocity w, then (a) the magnetic field produced at its centre will be \(B_0=\frac{\mu_0 q \omega}{2 \pi R}\)
5. The magnetic moment of the disc will be
   1. dM = (di) pix2
      • \(=\frac{\omega}{2 \pi} d q \pi x^2=\frac{\omega q}{R^2} x^3 d x\)
      • \(\Rightarrow \quad M=\int d M=\frac{\omega q}{R^2} \int_0^R x^3 d x\)
      • \(\Rightarrow \quad M=\frac{q \omega R^2}{4}\)
      • \(\Rightarrow \quad M=\frac{\mathrm{q} \omega \mathrm{R}^2}{4}\)

Question 21. A circular loop has a radius of 5 cm and it carries a current of 0.1 amp. Its magnetic moment is-

1. 1.32 × 10–4 amp. m2
2. 2.62 × 10–4 amp. m2
3. 5.25 × 10–4 amp. m2
4. 7.85 × 10–4 amp. m2

Answer: 4. 7.85 × 10–4 amp. m2

Question 22. The magnetic moment of a circular coil carrying current is-

1. Directly proportional to the length of the wire in the coil.
2. Inversely proportional to the length of the wire in the coil
3. Directly proportional to the square of the length of the wire in the coil
4. Inversely proportional to the square of the length of the wire in the coil.

Answer: 3. Directly proportional to the square of the length of the wire in the coil

Magnetic force on moving charge

When a charge q moves with velocity, in a magnetic field, then the magnetic force experienced by the moving charge is given by the following formula:

⇒ \(\vec{F}=q(\vec{v} \times \vec{B})\) Put q with sign.

⇒ \(\overrightarrow{\mathrm{v}}\): Instantaneous velocity

⇒ \(\overrightarrow{\mathrm{B}}\): Magnetic field at that point.

Note:

1. ⇒ \(\overrightarrow{\mathrm{F}} \perp \overrightarrow{\mathrm{V}} \text { and also } \overrightarrow{\mathrm{F}} \perp \overrightarrow{\mathrm{B}}\)
2. since \(\vec{F} \perp \vec{v}\) therefore power due to magnetic force on a charged particle is zero. (use the formula of power.
3. P= \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{V}}\) = for its proof).
4. Since the \(\overrightarrow{\mathrm{F}} \perp \overrightarrow{\mathrm{B}}\) work done by the magnetic force is zero in every part of the motion. The magnetic force cannot increase or decrease the speed (or kinetic energy) of a charged particle.
5. It can only change the direction of velocity.
6. On a stationary charged particle, the magnetic force is zero.
7. If \(\overrightarrow{\mathrm{V}} \| \overrightarrow{\mathrm{B}},\) then also magnetic force on charged particle is zero. It moves along a straight line if only a magnetic field is acting.

Solved Examples 

Example 10. A charged particle of mass 5 mg and charge q = +2uC has velocity \(\vec{v}=2 \hat{i}-3 \hat{j}+4 \hat{k}\). find out the magnetic force on the charged particle and its acceleration at this instant due to magnetic field \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \quad \overrightarrow{\mathrm{v}} \text { and } \overrightarrow{\mathrm{B}}\) are in m/s and web/m2 respectively.

Solution:

⇒ \(\vec{F}=q \vec{v} \times \vec{B}=2 \times 10^{-6}(2 \hat{i}-3 \hat{j} \times 4 \hat{k}) \times(3 \hat{j}-2 \hat{k})=2 \times 10^{-6}[-6 \hat{i}+4 \hat{j}+6 \hat{k}] N\)

By newton’s law \(\vec{a}=\frac{\vec{F}}{m}=\frac{2 \times 10^{-6}}{5 \times 10^{-6}}(-6 \hat{i}+4 \hat{j}+6 \hat{k})\)

⇒ \(=0.8(-3 \hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}^2\)

Example 11. A charged particle has acceleration \(\vec{a}=2 \hat{i}+x \hat{j}\) in a magnetic field \(\vec{B}=-3 \hat{i}+2 \hat{j}-4 \hat{k}\). Find the value of x.
Solution: Since \(\vec{F} \perp \vec{B}\) since \(\overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{B}}\) since \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{B}}=0\) ∴ \((2 \hat{i}+x \hat{j}) \cdot(-3 \hat{i}+2 \hat{j}-4 \hat{k})=0\)

-6+2x=0

x=3

Motion of charged particles under the effect of magnetic force

Particle released if v = 0 then fm = 0

∴ The particle will remain at rest

⇒ \(\vec{V} \| \vec{B} \text { here } \theta=0 \text { or } \theta=180^{\circ}\)

∴ Fm= 0

∴ \(\overrightarrow{\mathrm{a}}=0\)

∴ \(\vec{V}\) = const.

∴ The particle will move in a straight line with constant velocity

Initial velocity \(\overrightarrow{\mathrm{u}} \perp \overrightarrow{\mathrm{B}} \text { and } \overrightarrow{\mathrm{B}}\) = uniform

In this case, since B is in the z direction the magnetic force in the z-direction will be zero. (since \(\overrightarrow{F_m} \perp \vec{B}\)

∴ the particle will always move in the xy plane.

∴ velocity vector is always \(\perp \vec{B}\)

∴ \(F_m=\mathrm{quB}=\text { constant. now } \mathrm{quB}=\frac{\mathrm{mu}^2}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{\mathrm{mu}}{\mathrm{qB}}=\text { constant. }\)

The particle moves in a curved path whose radius of curvature is the same everywhere,
such a curve in a plane is only a circle

∴ The path of the particle is circular.

⇒ \(R=\frac{m u}{q B}=\frac{p}{q B}=\frac{\sqrt{2 m k}}{q B}\) Here p= linear momentum; k= kinetic energy

Now \(v=\omega R \Rightarrow \omega=\frac{q B}{m}=\frac{2 \pi}{T}=2 \pi f\)

Time period T = 2πm/qB frequency f = Bq/2πm

Note: ω, f, T is independent of velocity.

Example 12. A proton (p), α-particle and deuteron (D) are moving in circular paths with the same kinetic energies in the same magnetic field. Find the ratio of their radii and periods. (Neglect interaction between particles).
Solution: \(R=\frac{\sqrt{2 m K}}{q B}\)

∴ \(R_p: R_a: R_D=\frac{\sqrt{2 m K}}{q B}: \frac{\sqrt{2.4 m K}}{2 q B}: \frac{\sqrt{2.2 m K}}{q B}=1: 1: \sqrt{2}\) t=2πm/qB

∴ \(T_p: T_\alpha: T_D=\frac{2 \pi m}{q B}: \frac{2 \pi 4 m}{2 q B}: \frac{2 \pi 2 m}{q B}=1: 2: 2\)

Example 13. In the figure shown the magnetic field on the left of ‘PQ’ is zero and on the right of ‘PQ’ it is uniform. Find the time spent in the magnetic field.

Example 14. A uniform magnetic field of strength ‘B’ exists in a region of width ‘d’. A particle of charge ‘q’ and mass ‘m’ is shot perpendicularly (as shown in the figure) into the magnetic field. Find the time spent by the particle in the magnetic field if

⇒ \(d>\frac{\mathrm{mu}}{\mathrm{qB}}\)

⇒ \(d<\frac{m u}{q B}\)

Solution: \(d>\frac{\mathrm{mu}}{\mathrm{qB}}\)

∴ \(\mathrm{t}=\frac{\mathrm{T}}{2}=\frac{\pi \mathrm{m}}{q B}\)

⇒ \(\sin \theta=\frac{d}{R} \Rightarrow \theta=\sin ^{-1}\left(\frac{d}{R}\right)\) \(\omega t=\theta \Rightarrow t=\frac{m}{q B} \sin ^{-1}\left(\frac{d}{R}\right)\)

Question 15. What should be the speed of a charged particle so that it can’t collide with the upper wall? Also, find the coordinates of the point where the particle strikes the lower plate in the limiting case of velocity.

Solution: The path of the particle will be circular larger the velocity, the larger will be the radius. For particle not to s strike R < d.

Since \(\frac{m v}{q B}<d \quad \Rightarrow \quad v<\frac{q B d}{m} .\)

For limiting case \(v=\frac{q B d}{m}\)

R=d

Therefore coordinate = (–2d, 0, 0) l

Self Practice Problems 

Question 23. In a hydrogen atom, an e– moves in Bohr’s orbit of radius r = 5 x 10–11 m. and makes 1017 revolutions per second. The magnetic moment produced due to the orbital motion of the e– is-

1. 0.40π x 10–22 A-m2
2. 2.2π x 10–22 A-m2
3. 2π x 10–22 A-m2
4. None of these

Answer: 1. 0.40π x 10–22 A-m2

Question 24. A charged particle is moving with velocity v under the magnetic field B. The force acting on the particle will be maximum if-

1. v and B are in the same direction
2. v and B are in the opposite direction
3. v and B are perpendicular
4. v does not depend on the direction B.

Answer: 3. v and B are perpendicular

Helical path:

If the velocity of the charge is not perpendicular to the magnetic field, we can break the velocity into two components – V11 parallel to the field and \(V_{\perp}\) perpendicular to the field.

The components v11 remains unchanged as the force \(\mathrm{q} \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}\) is perpendicular to it. In the plane perpendicular to the field, the particle traces a circle of radius \(r=\frac{m v_{\perp}}{q B}\) as given by the equation. The resultant path is a helix.

Complete analysis:

Let a particle have an initial velocity in the plane of the paper and a constant and uniform magnetic field also in the plane of the paper.

The particle starts from point A1.

It completes its one revolution at A₂ 2nd revolution at A₃ and so on. X-axis is the tangent to the helix points.

A₁, A₂, A₃,……….all are on the x-axis. distance A₁ A₂ = A₃ A₄ = …………… = v cosθ. T = pitch where T = Period \(=\frac{\pi 2 m}{q B}\)

Let the particle’s initial position be (0,0,0) and v sinq in +y direction. Then
in x : Fx= 0, ax= 0, vx= constant = v cosθ, x = (v cosθ)t

In the y-z plane:

From the figure, it is clear that

⇒ \(\begin{aligned}
& y=R \sin \beta, v_y=v \sin \theta \cos \beta \\
& z=-(R-R \cos \beta) \\
& v_z=v \sin \theta \sin \beta
\end{aligned}\)

acceleration towards centre = (vsinθ)2/R = θ2R

∴ \(a_y=-\omega^2 R \sin \beta, a_z=-\omega^2 R \cos \beta\)

At any time: the position vector of the particle (or its displacement w.r.t. initial position)

⇒ \(\overrightarrow{\mathrm{r}}=x \hat{i}+y \hat{j}+z \hat{k}, x, y, z \text { already found }\)

Velocity \(\vec{v}=v_x \hat{i}+v_y \hat{j}+v_z \hat{k}, v_x, v_y, v_z \text { already found }\)

⇒ \(\vec{a}=a_x \hat{i}+a_y \hat{j}+a_z \hat{k}, a_x, a_y, a_z \text { already found }\)

Radius \(q(v \sin \theta) B=\frac{m(v \sin \theta)^2}{R} \quad \Rightarrow \quad R=\frac{m v \sin \theta}{q B}\)

⇒ \(\omega=\frac{v \sin \theta}{R}=\frac{q B}{m}=\frac{2 \pi}{T}=2 \pi f .\)

Charged Particle in \(\overrightarrow{\mathrm{E}} and \overrightarrow{\mathrm{B}}\)

When a charged particle moves with velocity in an electric field and magnetic field then. The net force experienced by it is given by the following equation.

⇒ \(\vec{F}=q \vec{E}+q(\vec{V} \times \vec{B})\)

Combine force is known as Lorentz force

image-

In the above situation, the particle passes undeviated but its velocity will change due to the electric field. Magnetic force on it = 0.

Case 1

⇒ \(\overrightarrow{\mathrm{E}} \| \overrightarrow{\mathrm{B}} \text { and uniform } \theta \neq 0 \text {, }\) (\(\text { } \vec{E} \text { and } \vec{B}\) are constant and uniform)

⇒ \(\text { in } x: F_x=q E, a_x=\frac{q E}{m}, v_x=v_0 \cos \theta+a_x t, x=v_0 t+\frac{1}{2} a_x t^2\)

in yz plane:

⇒ \(\begin{aligned}
& q v_0 \sin \theta B=m\left(v_0 \sin \theta\right)^2 / R \\
& \Rightarrow \quad R=\frac{m v_0 \sin \theta}{q B}, \quad \omega=\frac{v_0 \sin \theta}{R}=\frac{q B}{m}=\frac{2 \pi}{T}=2 \pi f
\end{aligned}\)

⇒ \(\vec{r}=\left\{\left(V_0 \cos \theta\right) t+\frac{1}{2} \frac{q E}{m} t^2\right\} \hat{i}+R \sin \omega t \hat{j}+(R-R \cos \omega t)(-\hat{k})\)

⇒ \(\vec{V}=\left(V_0 \cos \theta+\frac{q E}{m} t\right) \hat{i}+\left(V_0 \sin \theta\right) \cos \omega t \hat{j}+V_0 \sin \theta \sin \omega t(-\hat{k})\)

⇒ \(\vec{a}=\frac{q E}{m} \hat{i}+\omega^2 R[-\sin \beta \hat{j}-\cos \beta \hat{k}]\)

Magnetic force on a current-carrying wire:

B Suppose a conducting wire, carrying a current is placed in a magnetic field. Consider a small element dl of the wire (figure). The free electrons drift with a speed vd opposite to the direction of the current. The relation between the current i and the drift speed vd is

Here A is the area of the cross-section of the wire and n is the number of free electrons per unit volume. Each electron experiences an average (why average?) magnetic force.

⇒ \(\vec{f}=-e \vec{v}_d \times \vec{B}\)

The number of Free electrons in the small element is considered in nAdl. Thus, the magnetic force on the wire of length dl is

⇒ \(\mathrm{d} \overrightarrow{\mathrm{F}}=(\mathrm{nAd} \ell)\left(-\mathrm{e} \overrightarrow{\mathrm{v}}_{\mathrm{d}} \times \overrightarrow{\mathrm{B}}\right)\)

If we denote the length dl along the direction of the current, the above equation becomes

⇒ \(\begin{aligned}
& d \vec{F}=n A e v_d \vec{d} \ell \vec{B} \\
& d \vec{F}=i d \vec{\ell} \times \vec{B} .
\end{aligned}\)

using 1

The Quanity \(\text { id } \vec{\ell}\) is called a current element.

⇒ \(\overrightarrow{F_{\text {res }}}=\int \overrightarrow{\mathrm{dF}}=\int \mathrm{id} \vec{\ell} \times \overrightarrow{\mathrm{B}}=\mathrm{i} \int \overrightarrow{\mathrm{d} \ell} \times \overrightarrow{\mathrm{B}}\)

Since is the same at all points of the wire.

B If is uniform then \(\vec{F}_{\text {res }}=\mathrm{i}\left(\int \overrightarrow{\mathrm{d} \ell}\right) \times \overrightarrow{\mathrm{B}}\)

⇒ \(\overrightarrow{F_{\text {res }}}=\mathrm{i} \overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{B}}\)

Here \(\overrightarrow{\mathrm{L}}=\int \mathrm{d} \vec{\ell}\) = vector length of the wire = vector connecting the endpoints of the wire.

The direction of magnetic force is perpendicular to the plane of \(\overrightarrow{\mathrm{I}} \text { and } \overrightarrow{\mathrm{B}}\) according to right-hand screw rule. The following two rules are used in determining the direction of the magnetic force.

Right-hand palm rule: If the right hand and the palm are stretched such that the thumb points in the direction of current and the stretched fingers in the direction of the magnetic field, then the force on the conductor will be perpendicular to the palm in the outward direction.

Fleming left-hand rule: If the thumb, forefinger, and central finger of the left hand are stretched such that the first finger points in the direction of the magnetic field and the central finger in the direction of current, then the thumb will point in the direction of force acting on the conductor.

Note: f a current loop of any shape is placed in a uniform

⇒ \(\left.\overrightarrow{\mathrm{B}} \text { then } \overrightarrow{\mathrm{F}_{\text {res }}}\right)_{\text {magnetic }}\) on it=0 (since \(\overrightarrow{\mathrm{L}}=0\)

Point of application of magnetic force:

On a straight current-carrying wire the magnetic force in a uniform magnetic field can be assumed to be acting at its midpoint.

This can be used for the calculation of torque.

Example 16. A wire is bent in the form of an equilateral triangle PQR of side 10 cm and carries a current of 5.0 A. It is placed in a magnetic field B of magnitude 2.0 T directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle.
Solution: Suppose the field and the current have directions as shown in the figure. The force on PQ is

⇒ \(\vec{F}_1=\overrightarrow{i \ell} \times \vec{B} \quad \text { or, } \quad F_1=5.0 \mathrm{~A} \times 10 \mathrm{~cm} \times 2.0 \mathrm{~T}=1.0 \mathrm{~N}\)

The rule of vector product shows that the force F 1 is perpendicular to PQ and is directed towards the inside of the triangle. The forces \(\vec{F}_2 \text { and } \vec{F}_3\) on QR and RP can also be obtained similarly. Both the forces are 1.0 N
directed perpendicularly to the respective sides and towards the inside of the triangle.

The three forces \(\vec{F}_1, \vec{F}_2 \text { and } \vec{F}_3\) will have zero resultant so that there is no net magnetic force on the triangle. This result can be generalized. Any closed current loop, placed in a homogeneous magnetic field, does not experience a net magnetic force.

Example 17. Two long wires, carrying currents i1 and i2, are placed perpendicular to each other in such a way that they just avoid contact. Find the magnetic force on a small length dl of the second wire situated at a distance from the first wire.

Solution: The situation is shown in the figure. The magnetic field at the site of d, due to the first wire is, \(\mathrm{B}=\frac{\mu_0 i_1}{2 \pi \ell}\)

This field is perpendicular to the plane of the figure going into it. The magnetic force on the length dl is,

⇒ \(\mathrm{dF}=\mathrm{i}_2 \mathrm{~d} \ell \mathrm{B} \sin 90^{\circ}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \mathrm{~d} \ell}{2 \pi \ell}\)

This force is parallel to the current i1.

Example 18. The figure shows two long metal rails placed horizontally and parallel to each other at a separation l. A uniform magnetic field B exists in the vertically downward direction. A wire of mass m can slide on the rails. The rails are connected to a constant current source which drives a current i in the circuit. The friction coefficient between the rails and the wire is u.

1. What soluble minimum value can prevent the wire from sliding on the rails?
2. Describe the motion of the wire if the value of u is half the value found in the previous part.

The force on the wire due to the magnetic field is

⇒ \(\overrightarrow{\mathrm{F}}=\mathrm{i} \vec{\ell} \times \overrightarrow{\mathrm{B}} \quad \text { or, } \quad \mathrm{F}=\mathrm{i} \ell \mathrm{B}\)

It acts towards the right in the given figure. If the wire does not slide on the rails, the force of friction by the rails should be equal to F. If u0 is the minimum coefficient of friction which can prevent sliding, this force is also equal to u0 mg. Thus,

⇒ \(\mu_0 \mathrm{mg}=\mathrm{i} \ell \mathrm{B} \quad \text { or, } \quad \mu_0=\frac{\mathrm{i} \ell \mathrm{B}}{\mathrm{mg}}\)

If the friction coefficient is \(\mu=\frac{\mu_0}{2}=\frac{i \ell B}{2 \mathrm{mg}}\) the wire will slide towards right. The frictional force by the rails is

⇒ \(f=\mu \mathrm{mg}=\frac{\mathrm{i} \ell \mathrm{B}}{2}\) towards left.

The resultant force is \(\mathrm{i} \ell \mathrm{B}-\frac{\mathrm{i} / \mathrm{B}}{2}=\frac{\mathrm{i} / \mathrm{B}}{2}\) towards right. The acceleration will be \(\mathrm{a}=\frac{\mathrm{i} \ell \mathrm{B}}{2 \mathrm{~m}}\) The
the wire will slide towards the right with this acceleration.

Example 19. In the figure shown a semicircular wire is placed in a uniform directed toward the right. Find the resultant magnetic force and torque on it.

Solution: The wire is equivalent to forces on individual parts marked in the figure by and . By symmetry there will be a pair of forces forming couples.

⇒ \(\tau=\int_0^{\pi / 2} i(R d \theta) B \sin (90-\theta) \cdot 2 R \cos \theta\)

⇒ \(\tau=\frac{i \pi R^2}{2} B \quad \Rightarrow \quad \vec{\tau}=\frac{i \pi R^2}{2} B(-\hat{j})\)

Example 20. Find the resultant magnetic force and torque on the loop

Solution: \(\overrightarrow{F_{\text {res }}}=0 \text {, }\) (since loop) and \(\vec{\tau}=i \pi R^2 B(-\hat{j})\) using the above method.

Example 21. In the figure shown find the resultant magnetic force and torque about ‘C’, and ‘P’.

Solution:

⇒ \(\begin{aligned}
& \mathrm{F}=\mathrm{I} \ell_{\mathrm{eq}} \mathrm{B} \\
& \overrightarrow{\mathrm{F}}_{\text {nett }}=\mathrm{I} .2 \mathrm{R} . \mathrm{B}
\end{aligned}\)

since the wire is equivalent to

Force on each element is radially outward: tc = 0 point about

 

Example 22. Prove that magnetic force per unit length on each of the infinitely long w ire due to each other is \(\mu_0 I_1 I_2 / 2 \pi d\). Here it is attractive also.

Solution:

on (2), B due to 1= \(=\frac{\mu_0 \mathrm{I}_1}{2 \pi \mathrm{d}} \otimes\)

therefore F on (2) on 1 m length

⇒ \(=I_2 \cdot \frac{\mu_0 I_1}{2 \pi d} \cdot 1\) towards left it is attractive

⇒ \(=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}\) (hence proved)

Similarly on the other wire also.

Note:

Definition of ampere (the fundamental unit of current) using the above formula.

If I1 = I2 = 1A, d = 1m then F = 2 × 10–7 N

“When two very long wires carrying equal currents and separated by 1m distance exert on
each other a magnetic force of 2 × 10–7 N on 1m length then the current is 1 ampere.”

The above formula can also be applied if one wire is infinitely long and the other is of finite length. In this case, the force per unit length on each wire will not be the same.

Force per unit length on \(P Q=\frac{\mu_0 I_1 I_2}{2 \pi d}\) (attractive)

If the currents are in the opposite direction then the magnetic force on the wires will be repulsive.

Solved Examples

Example 23. Find the magnetic force on the loop ‘PQRS’ due to the loop wire.
Solution: \(\mathrm{F}_{\text {ras }}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{a}} a(-\hat{\mathrm{i}})+\frac{\mu_0 \mathrm{I}_4 \mathrm{I}_2}{2 \pi(2 \mathrm{a})} \mathrm{a}(\hat{\mathrm{i}})=\frac{\mu_0 \mathrm{I}_4 \mathrm{I}_2}{4 \pi}(-\hat{\mathrm{i}})\)

Example 24. A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and. CD. A steady current I is flowing in the loop. The angle made by AB and CD at the origin O is 30º. Another straight thin wire with a steady current I1 flowing out of the plane of the paper is kept at the origin.

The magnitude of the magnetic field due to the loop ABCD at the origin (O) is:

1. latex]\frac{\mu_0 I(b-a)}{24 a b}[/latex]
2. \(\frac{\mu_0 \mathrm{I}}{4 \pi}\left[\frac{\mathrm{b}-\mathrm{a}}{\mathrm{ab}}\right]\)
3. \(\frac{\mu_0 I}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]\)
4. Zero

Solution: Magnetic field due to loop ABCD

⇒ \(=\frac{\mu_0 1}{4 \pi}\left(\frac{\pi}{6}\right) \times\left[\frac{1}{a}-\frac{1}{b}\right]=\frac{\mu_0 1}{24}\left[\frac{b-a}{a b}\right]\)

Due to the presence of the current I1 at the origin:

1. The forces on AD and BC are zero.
2. The magnitude of the net force on the loop is given by \(\frac{\mu_0 I_1 I}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]\)
3. The magnitude of the net force on the loop is given by \(\frac{\mu_0 \mathrm{II}_1}{24 \mathrm{ab}}(\mathrm{b}-\mathrm{a}) \text {. }\)
4. The forces on AB and DC are zero.

Solution: \(\vec{F}=\mathrm{i}(\vec{\ell} \times \vec{B})\)

The magnetic field due to I1 is parallel to AD and BC. So that force On AD and BC is zero.

Self-practice problems

Question 25. The force on a conductor of length l placed in a magnetic field of magnitude B and carrying in current I is given by (θ is the angle which the conductor makes with the direction of B)

1. I l B sin θ
2. I2lB2 sin θ
3. IlB Cosθ
4. \(\frac{\mathrm{I}^2 \ell}{\mathrm{B}} \sin \theta\)

Answer: 1. I l B sin θ

Question 26. A charge ‘q’ moves in a region where an electric field and magnetic field both exist, then the net force on it-

1. \(q(\vec{v} \times \vec{B})\)
2. \(q \vec{E}+q(\vec{v} \times \vec{B})\)
3. \(q \vec{E}+q(\vec{B} \times \vec{v})\)
4. \(q \vec{B}+q(\vec{E} \times \vec{v})\)

Answer: 2. \(q \vec{E}+q(\vec{v} \times \vec{B})\)

Question 27. An electron moves with speed 2 × 105 m/s along the positive x-direction in the presence of a magnetic \(\vec{B}=\hat{i}+4 \hat{j}-3 \hat{k}\) induction (in tesla). The magnitude of the force experience by the electron in newtons is- (charge on the electron = 1.6 × 10–19 C)

1. 1.18 × 10–13
2. 1.28 × 10–13
3. 1.6 × 10–13
4. 1.72 × 10–13

Answer: 3. 1.6 × 10–13

Question 28. An electron (charge q coulomb) enters a magnetic field of H weber/m 2 with a velocity of v m/s in the same

1. Direction as that of the field. The force on its electron is-
2. Hqv newtons in the direction of the magnetic field
3. Hqv dynes in the direction of the magnetic field
4. Hqv newtons at a right angle to the direction of the magnetic field
5. Zero

Answer: 4. Zero

Question 29. A long wire A carries a current of 10 amp. Another long wire B, which is parallel to A and separated by 0.1 m from A, carries a current of 5 amp. in the opposite direction to that in A. What is the magnitude and nature of the force experience per unit length of B- \(\left[\mu_0=4 \pi \times 10^{-1}\right? \text { W/amp-m] }\)

1. Repulsive force of 10–4 N/m
2. Attractive force of 10–4 N/m
3. Repulsive force of 2π x 10–5 N/m
4. Attractive force of 2π × 10–5 N/m

Answer: 1. Repulsive force of 10–4 N/m.

Torque on a current loop:

When a current-carrying coil is placed in a uniform magnetic field the net force on it is always zero.

However, as its different parts experience forces in different directions so the loop may experience a torque (or couple) depending on the orientation of the loop and the axis of rotation. For this, consider a rectangular coil in a uniform field B which is free to rotate about a vertical axis PQ and normal to the plane of the coil making an angle  with the field direction

The arms AB and CD will experience forces B(NI)b vertically up and down respectively. These two forces together will give zero net force and zero torque (as are collinear with the axis of rotation), so will not affect the motion of the coil.

Now the forces on the arms AC and BD will be BINL in the direction out of the page and into the page respectively, resulting in zero net force, but an anticlockwise couple of value \(\tau=F \times A r m=B I N L \times(b \sin \theta)\)

ie. τ = BIA sinθ with A = NLb ………….(1)

M A=  Now treating the current–carrying coil as a dipole of the moment \(\overrightarrow{\mathrm{M}}=\mathrm{I} \overrightarrow{\mathrm{A}}\) Eqn.

Can be written in vector form as \(\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}\) \(\text { [with } \vec{M}=I \vec{A}=\text { NIA } \vec{n}\)

This is the required result and from this, it is clear that

Torque will be minimum (= 0) when sinθ = min = θ, i.e., θ = 0º, i.e. 180º i.e., the plane of the coil is perpendicular to the magnetic field i.e. normal to the coil is collinear with the field.

Torque will be maximum (= BINA) when sin = max = 1, i.e., θ = 90º i.e. the plane of the coil is parallel to the field i.e. normal to the coil is perpendicular to the field.

By analogy with dielectric or magnetic dipole in a field, in case of current–carrying in a field.

\(\mathrm{U}=\overrightarrow{\mathrm{M}} \bullet \overrightarrow{\mathrm{B}} \quad \text { with } \quad \mathrm{F}=\frac{\mathrm{dU}}{\mathrm{dr}}\)

and W = MB(1 – cosθ)
The values of U and W for different orientations of the coil in the field.

Instruments such as electric motors, moving coil galvanometers tangent galvanometers etc. are based on the fact that a current–carrying coil in a uniform magnetic field experiences a torque (or couple).

Example 25: A bar magnet having a magnetic moment of 2 × 10 4 JT–1 is free to rotate in a horizontal plane. A horizontal magnetic field B= 6 × 10–4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is
Solution: The work done in rotating a magnetic dipole against the torque acting on it, when placed in a magnetic field is stored inside it in the form of potential energy. When magnetic dipole is rotated from initial position θ = θ1 to final position θ = θ2, then work done = MB(cos θ1– cos θ2)

⇒ \(=\operatorname{MB}\left(1-\frac{1}{2}\right)=\frac{2 \times 10^4 \times 6 \times 10^{-4}}{2}=6 \mathrm{~J}\)

Self-practice Problems

Due to the flow of current in a circular loop of radius R, the magnetic induction produced at the center of the loop is B. The magnetic moment of the loop is [u0 = Permeability of vacuum]

1. \(\frac{\mathrm{BR}^3}{2 \pi \mu_0}\)
2. \(\frac{2 \pi \mathrm{BR}^3}{\mu_0}\)
3. \(\frac{\mathrm{BR}^2}{2 \pi \mu_0}\)
4. \(\frac{2 \pi \mathrm{BR}^2}{\mu_0}\)

Answer: 2. \(\frac{2 \pi \mathrm{BR}^3}{\mu_0}\)

Question 31. A current i flows in a circular coil of radius r. If the coil is placed in a uniform magnetic field B with its plane parallel to the field magnitude of torque act on the coil is-

1. Zero
2. 2πriB
3. πr2iB
4. 2π2iB

Answer: 3. πr2iB

Question 32. To double the torque acting on a rectangular coil of n turns, when placed in a magnetic field-

1. The area of the coil and the magnetic induction should be doubled.
2. The area and current through the coil should be doubled.
3. Only the area of the coil should be doubled.
4. Some turns are to be halved.

Answer: 3. Only the area of the coil should be doubled.

Question 33. An arbitrarily shaped closed coil is made of a wire of length L and a current ampere is flowing in it. If B the plane of the coil is perpendicular to the magnetic field \(\vec{B}\) The force on the coil is

1. Zero
2. IBl
3. IBL
4. \(\frac{1}{2} \text { IBL }\)

Answer: 1. Zero

Question 34. A current-carrying loop is placed in a uniform magnetic field in four different orientations, 1,2, 3, and 4 arranged in the decreasing order of potential energy

1. 1>3>2>4
2. 1>2>3>4
3. 1>4>2>3
4. 3>4>1>2

Answer: 3. 1>4>2>3

Example: 26 (Read the following passage and answer the questions. They have only one correct option) In the given figure of a cyclotron, show the particle source S and the dees. A uniform magnetic field is directed up from the plane of the page. Circulating protons spiral outward within the hollow dees, gaining energy every time they cross the gap between the dees.

Suppose that a proton, injected by source S at the center of the cyclotron in Fig., initially moves toward a negatively charged dee. It will accelerate toward this dee and enter it. Once inside, it is shielded from the electric field by the copper walls of the dee; that is the electric field does not enter the dee.

The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in a circular path whose radius, which depends on its speed, is given by

⇒ \(\text { Eq. } r=\frac{m v}{q B}\)

Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed. Thus, the proton again faces a negatively charged dee and is again accelerated.

Thus, the proton again faces a negatively charged dee and is again accelerated. This process continues, the circulating proton always being in step, with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system. There a deflector plate sends it out through a portal.

The key to the operation of the cyclotron is that the frequency f at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency fosc of the electrical oscillator, or \(\mathrm{f}=\mathrm{f}_{\text {oss }} \text { (resonance condition). }\)

This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency fosc that is equal to the natural frequency f at which the proton circulates in the magnetic field.

Combining Eq. 1 and 2 allows us to write the resonance condition as \(\mathrm{qB}=2 \pi \mathrm{mf}_{\mathrm{osc}}\)

For the proton, q and m are fixed. The oscillator (we assume) is designed to work at a single fixed frequency fosc. We then “tune” the cyclotron by varying B until eq. 3 is satisfied and then many protons circulate through the magnetic field, to emerge as a beam.

Ratio of the radius of successive semi circular path

1. \(\sqrt{1}: \sqrt{2}: \sqrt{3}: \sqrt{4}\ldots \ldots \ldots \ldots\)
2. \(\sqrt{1}: \sqrt{3}: \sqrt{5}\ldots \ldots \ldots \ldots\)
3. \(\sqrt{2}: \sqrt{4}: \sqrt{6} \ldots \ldots \ldots \ldots\)
4. \(1: 2: 3\ldots \ldots \ldots \ldots\)

Solution: When the charge is accelerated by an electric field it gains energy for the first time \(K E_1=\frac{q V}{2}\)

For second time \(\mathrm{KE}_2=\frac{3}{2} \mathrm{qV}\)

For third time ,\(K E_3=\frac{5}{2} q V\)

Hence The ratio of radii are

⇒ \(r_1: r_2: r_3: \ldots \ldots \ldots \ldots: \frac{\sqrt{2 m \frac{q v}{2}}}{q B}: \frac{\sqrt{2 m \frac{3}{2} q v}}{q B}: \ldots \ldots . .\)

⇒ \(r_1: r_2: r_3 \ldots \ldots \ldots:: \sqrt{1}: \sqrt{3}: \sqrt{5}\)

Change in kinetic energy of charged particle after every period is:

1. 2qV
2. qV
3. 3qV
4. None of these

Solution: In one full cycle it gets accelerated two times so change in KE = 2 qV.

If q/m for a charged particle is 106, the frequency of applied AC is 106 Hz. Then the applied magnetic field is:

(1) 2π tesla (2) π tesla (3) 2 tesla (4) can not be defined

Solution: \(f=\frac{q B}{2 \pi m} \quad \Rightarrow 10^5=\frac{10^{\circ} B}{2 \pi} \quad \Rightarrow 2 \pi T .\)

Distance traveled in each period is in the ratio of:

1. \(\sqrt{1}+\sqrt{3}: \sqrt{5}+\sqrt{7}: \sqrt{9}+\sqrt{11}\)
2. \(\sqrt{2}+\sqrt{3}: \sqrt{4}+\sqrt{5}: \sqrt{6}+\sqrt{7}\)
3. \(\sqrt{1}: \sqrt{2}: \sqrt{3}\)
4. \(\sqrt{2}: \sqrt{3}: \sqrt{4}\)

Solution: Distance traveled by particle in one time period:

⇒ \(\pi\left(r_1+r_2\right): \pi\left(r_3+r_4\right): \pi\left(r_5+r_6\right)\)

⇒ \(\frac{\sqrt{2 m \frac{q V}{2}}}{q B}+\frac{\sqrt{2 m \frac{3 q V}{2}}}{q B}: \frac{\sqrt{2 m \frac{5 q V}{2}}}{q B}+\frac{\sqrt{2 m \frac{7 q V}{2}}}{q B}: \frac{\sqrt{2 m \frac{9 q V}{2}}}{q B}+\frac{\sqrt{2 m \frac{11 q V}{2}}}{q B} \ldots \ldots\)

⇒ \(S_1: S_2: S_3 \ldots \ldots \ldots \ldots \ldots:(\sqrt{1}+\sqrt{3}):(\sqrt{5}+\sqrt{7}):(\sqrt{9}+\sqrt{11})\)

For a given charge particle a cyclotron can be “tuned” by :

1. Changing applied a.c. Voltage only
2. Changing applied a.c. Voltage and magnetic field both
3. Changing the applied magnetic field only
4. By changing the frequency of applied a.c.

Solution: The frequency of A.C. depends on charge and mass only so it can be tuned by magnetic field only.

Terrestrial Magnetism (Earth’s Magnetism):

Introduction :

The idea that the earth is magnetized was first suggested towards the end of the sixteenth century by Dr. William Gilbert. The origin of Earth’s magnetism is still a matter of conjecture among scientists but it is agreed upon that the Earth behaves as a magnetic dipole inclined at a small angle (11.5º) to the Earth’s axis of rotation with its south pole pointing north.

The lines of force of the earth’s magnetic field are shown in the figure which is parallel to the earth’s surface near the equator and perpendicular to it near the poles.

While discussing the magnetism of the earth one should keep in mind that:

The magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet, i.e., it is a plane that contains the place and the magnetic axis.

The geographical meridian at a place is a vertical plane that passes through the line joining the geographical north and south, i.e., it is a plane that contains the place and earth’s axis of rotation, i.e., the geographical axis.

The magnetic Equator is a great circle (a circle with the center at the earth’s center) on the earth’s surface which is perpendicular to the magnetic axis. The magnetic equator passing through Trivandrum in South India divides the earth into two hemispheres.

The hemisphere containing the south polarity of the earth’s magnetism is called the northern hemisphere (NHS) while the other is the southern hemisphere (SHS).

The magnetic field of the earth is not constant and changes irregularly from place to place on the surface of the earth and even at a given place it varies with time too.

Elements of the Earth’s Magnetism:

The magnetism of Earth is completely specified by the following three parameters called elements of Earth’s magnetism:

Variation or Declination θ: At a given place the angle between the geographical meridian and the magnetic meridian is called declination, i.e., at a given place it is the angle between the geographical north-south direction and the direction indicated by a magnetic compass needle,

Declination at a place is expressed at θº E or θº W depending upon whether the north pole of the compass needle lies to the east (right) or to the west (left) of the geographical north-south direction. The declination at London is 10ºW means that at London the north pole of a compass needle points 10ºW, i.e., left of the geographical north.

Inclination or Angle of Dip Φ: It is the angle at which the direction of the resultant intensity of the earth’s magnetic field subtends with a horizontal line in the magnetic meridian at the given place.

It is the angle at which the axis of a freely suspended magnet (up or down) subtends with the horizontal in the magnetic meridian at a given place.

Here, it is worth noting that as the northern hemisphere contains the south polarity of earth’s magnetism, in it the north pole of a freely suspended magnet (or pivoted compass needle) will dip downwards, i.e., towards the earth while the opposite will take place in the southern hemisphere.

The angle of dip at a place is measured by the instrument called Dip-Circle in which a magnetic needle is free to rotate in a vertical plane which can be set in any vertical direction. The angle of dip at Delhi is 42º.

Horizontal Component of Earth’s Magnetic Field BH: At a given place it is defined as the component of Earth’s magnetic field along the horizontal in the magnetic meridian. It is represented by BH and is measured with the help of a vibration or deflection magnetometer.

At Delhi, the horizontal component of the earth’s magnetic field is 35 μT, i.e., 0.35 G. If at a place the magnetic field of the earth is BI and the angle of dip Φ, then under figure (a).

and \(\begin{aligned}
& \mathrm{B}_{\mathrm{H}}=\mathrm{B}_1 \cos \phi \\
& \mathrm{B}_{\mathrm{v}}=\mathrm{B}_1 \sin \phi
\end{aligned}\)

So that, \(\tan \phi=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}}\)

And \(I=\sqrt{B_H^2+B_v^2}\) …..(2)

Self Practice Problems

Question 35. At a place, the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are 0.3 and 0.6 oersted respectively. The value of the angle of dip at this place will be

1. 60°
2. 45°
3. 30°
4. 0°

Answer: 1. 60°

Question 36. The angle of dip at a place on the earth gives-

1. The horizontal component of the earth’s magnetic field.
2. The location of the geographic meridian.
3. The vertical component of the earth’s field.
4. The direction of the earth’s magnetic field.

Answer: 4. The direction of the earth’s magnetic field.

Question 37. A bar magnet is placed north-south with its north pole due north. The points of zero magnetic field will be in which direction from the center of the magnet-

1. North and south
2. East and west
3. Northeast and southwest
4. Northwest and southeast

Answer: 2. East and west

Question 38. When the N-pole of a bar magnet points towards the south and the S-pole towards the north, the null points are at the-

1. Magnetic axis
2. Magnetic centre
3. Perpendicular divider of magnetic axis
4. N and s-pole

Answer: 3. Perpendicular divider of the magnetic axis

Magnetic Substances And Their Properties:

Classification of substances according to their magnetic behavior:

All substances show magnetic properties. An iron nail brought near a pole of a bar magnet is strongly attracted by it and sticks to it, Similar is the behavior of steel, cobalt, and nickel. Such substances are called ‘ferromagnetic ‘substances.

Some substances are only weakly attracted by a magnet, while some are repelled by it. They are called ‘paramagnetic’ and ‘diamagnetic ‘substances respectively. All substances, solids, liquids, and gases, fall into one or other of these classes.

Diamagnetic substance: Some substance, when placed in a magnetic field, are feebly magnetized opposite to the direction of the magnetizing field. These substances when brought close to a pole of a powerful magnet, are somewhat repelled away from the magnet. They are called ‘diamagnetic’ substances and their magnetism is called ‘diamagnetism’.

Examples of diamagnetic substances are bismuth, zinc, copper, silver, gold, lead, water, mercury, sodium chloride, nitrogen, hydrogen, etc.

Paramagnetic substances: Some substance when placed in a magnetic field, are feebly magnetized in the direction of the magnetizing field. This substance, when brought close to a pole of a powerful magnet, is attracted towards the magnet. These are called ‘paramagnetic’ substances and their magnetism is called ‘paramagnetism’.

Ferromagnetic substances: Some substance, when placed in a magnetic field, are strongly magnetized in the direction of the magnetizing field. They are attracted fast towards a magnet when brought close to either of the poles of the magnet. These are called ‘ferromagnetic’ substances and their magnetism is called ‘ferromagnetism’

Some important terms used in magnetism:

Magnetic induction:

When a piece of any substance is placed in an external magnetic field, the substance becomes magnetized. The magnetism so produced in the substance is called ‘induced magnetism’ and this phenomenon is called ‘magnetic induction’.

The number of magnetic lines of induction inside a magnetized substance crossing the unit area normal to their direction is called the magnitude of magnetic induction, or magnetic flux density, inside the substance. It is denoted by B.

Magnetic induction is a vector whose direction at any point is the direction of the magnetic line of induction at that point. The SI unit of magnetic induction is the tesla (T) or Weber/meter2 (Wb-m–2) or Newton/(amper-meter) (NA–1m–1). The CGS unit is ‘gauss’.

Intensity of magnetization (I):

The intensity of magnetization, or simply magnetization of a magnetized substance represents the extent to which the substance is magnetized.

It is defined as the magnetic moment per unit volume of the magnetized substance. It is denoted by I. Its SI unit is ampere/meter (Am–1). Numerically. I= M/V

In the case of a bar magnet, if m is the pole strength of the magnet, 2l is its magnetic length, and its area of cross-section, then

⇒ \(\mathrm{I}=\frac{\mathrm{M}}{\mathrm{V}}=\frac{\mathrm{m} \times 2 \ell}{\mathrm{a} \times 2 \ell}=\frac{\mathrm{m}}{\mathrm{a}}\)

Thus, magnetization may also be defined as pole strength per unit area of cross-section.

Magnetic Intensity or Magnetic Field strength:

When a substance is placed in an external magnetic field, it becomes magnetized. The actual magnetic field inside the substance is the sum of the external field due to its magnetization. The (H) capability of the magnetizing field to magnetize the substance is expressed through a vector, called the ‘magnetic intensity’ of the field. It is defined through the vector relation.

⇒ \((\vec{H})=\frac{B}{\mu_0}-(\overrightarrow{\mathrm{I}}) \text {. }\)

Where is magnetic field induction inside the substance and is the intensity of magnetization? µ 0 is the permeability of space.

The SI unit of is the same as of, that is, ampere/meter (Am–1). The C.G .S. unit is ‘oersted’.

Magnetic permeability (µ):

The magnetic permeability of a substance is a measure of its conduction of magnetic lines of force (B) through it. It is defined as the ratio of the magnetic induction inside the magnetized substance to the magnetic intensity of the magnetizing field, that is,

⇒ \(\mu=\frac{\vec{B}}{\vec{H}} .\)

Numerically, µ = B/H.

ts SI unit is Weber/amper-metre) (Wb A–1 m –1) or \(\frac{\text { Newton }}{\text { Ampere }^2}\left(\mathrm{NA}^{-2}\right) \text {. }^2\)

Relative magnetic permeability (µr): The relative magnetic permeability of a substance is the ratio of the magnetic permeability µ of the substance to the permeability of free space µ 0, that is, \(\mu_r=\frac{\mu}{\mu_0} .\)

It is a dimensionless quantity and is equal to 1 for vacuum (by definition). Alternatively, the relative permeability of a substance is defined as the ratio of the magnetic flux density B in the substance when placed in a magnetic field and the flux density B 0 in a vacuum in the same field, that is,

⇒ \(\mu_r=\frac{B}{B_0} .\)

We can classify substances in terms of µr:

µr < 1 (diamagnetic)
µr > 1 (paramagnetic)
µr > > 1 (ferromagnetic)

Magnetic susceptibility (xm):

It is a measure of how easily a substance is magnetized in a magnetizing field. For paramagnetic and (H) diamagnetic substances, the magnetization is directly proportional to the magnetic intensity of the magnetizing field. That is.

⇒ \(\overrightarrow{\mathrm{I}}=\chi_{\mathrm{m}}(\overrightarrow{\mathrm{H}})\)

The constant xm is called the ‘magnetic susceptibility of the substance. It may be defined as the ratio of the intensity of magnetization to the magnetic intensity of the magnetizing field, that is, \(\chi_m=\frac{I}{H} \text {. }\)

It is a pure number because I and H have the same unit). Its value for vacuum is zero as there can be no magnetization in a vacuum. We can classify substances in terms of xm. Substances with positive values of xm are paramagnetic and those with negative values of xm are diamagnetic. For ferromagnetic substances, xm is positive and very (H) x large. However, for them, is not accurately proportional to, and so xm is not strictly constant.

Relation between Relative permeability (µ r) and magnetic susceptibility (xm): When a substance is placed in a magnetizing field, it becomes magnetized. The total magnetic flux density B within the substance is the flux density that would have been produced by the magnetizing field in vacuum plus the flux density due to the magnetization of the substance. If I am the intensity of magnetization of the substance, then, by definition, the magnetic intensity of the magnetizing field is given by –

⇒ \(H=\frac{B}{\mu_0}-I\)

Or \(\text { But } I=\chi_m H \text {, where } \chi_m\) is the susceptibility of the substance. B = µ 0 (H + I).
But I = xm H, where µ is the permeability of the substance.

⇒ \(\mu=\mu_0\left(1+\chi_m\right) \text {. }\)

or \(\frac{\mu}{\mu_0}=1+\chi_m\)

Properties of dia, para, and ferromagnetic substance: Diamagnetic substance: These substances are feebly repelled by a magnet. When placed in a magnetizing field, they are feebly magnetized in a direction opposite to that of the field.

Thus, the susceptibility Im of a diamagnetic substance is negative: Further, the flux density in a diamagnetic substance placed in a magnetizing field is slightly less than in the free space. Thus, the relative permeability µ r is less than 1.

Diamagnetic substances show the following properties.

When a rod of diamagnetic material is suspended freely between two magnetic poles, then its axis becomes perpendicular to the magnetic field.

In a non-uniform magnetic field, a diamagnetic substance tends to move from the stronger to the weaker part of the field.

If a diamagnetic solution is poured into a U-tube and one arm of this U-tube is placed between the poles of a strong magnet, the level of the solution in that arm is depressed.

A diamagnetic gas when allowed to ascend in between the poles of a magnet spreads across the field.

The susceptibility of a diamagnetic substance is independent of temperature. Paramagnetic substance: These substances are feebly attracted by a magnet. When placed in a magnetizing field, they are feebly magnetized in the direction of the field. T

Thus, they have a positive susceptibility xm The relative permeability µ r for paramagnetics is slightly greater than 1:

When a rod of paramagnetic material is suspended freely between two magnetic poles, then its axis becomes parallel to the magnetic field The poles produced at the ends of the rod are opposite to the nearer magnetic poles.

In a non-uniform magnetic field, the paramagnetic substances tend to move from the weaker to the stronger part of the magnetic field.

If a paramagnetic solution is poured into a U-tube and one arm of the U-tube is placed between two strong poles, the level of the solution in that arm rises.

A paramagnetic gas when allowed to ascend between the pole-pieces of a magnet, spreads along the field.

The susceptibility of a paramagnetic substance varies inversely to the kelvin temperature of the substance, that is, \(\chi_m \propto \frac{1}{T}\)

This is known as Curie’s law.

Ferromagnetic substances: These substances which are strongly attracted by a magnet, show all the properties of a paramagnetic substance to a much higher degree.

For example, they are strongly magnetized in the relatively weak magnetizing field in the same direction as the field. They have relative permeabilities of the order of hundreds and thousands. Similarly, the susceptibilities of ferromagnetic have large positive values.

Curie temperature: Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance “suddenly” disappears and the substance becomes paramagnetic.

The temperature above which a ferromagnetic substance becomes paramagnetic is called the’ Curie temperature’ of the substance. The curie temperature of iron is 770ºC and that of nickel is 358ºC.

Explanation of Dia-, para- and ferromagnetism based on the atomic model of magnetism:

The diamagnetic, paramagnetic, and ferromagnetic behavior of substances can be explained based on the atomic model, we know that matter is made up of atoms.

Each atom of any substance has a positively charged nucleus at its center around which electrons revolve in various discrete orbits.

Each revolving electron is equivalent to a tiny current loop (or magnetic dipole) and gives a dipole moment to the atom.

Besides this, each electron “spins” about its axis and this spin also produces a magnetic dipole moment. However, most of the magnetic moment of the atom is produced by electron spin, the contribution of the orbital revolution is very small.

Explanation of Diamagnetism: The property of diamagnetism is generally found is those substances whose atoms, or molecules, have an “even’ number of electrons that form pairs. The direction of the spin of one electron is opposite to that of the other. So, the magnetic moment of one electron is neutralized by that of the other.

As such, the net magnetic moment of an atom of a diamagnetic substance is zero. Diamagnetism is temperature-independent.

Explanation of paramagnetism:

The property of paramagnetism is found in a substance whose atoms, or molecules, have an excess of electrons spinning in the same direction. Hence atoms of paramagnetic substance have a permanent magnetic moment and behave like tiny bar-magnets.

Even then the paramagnetic substances do not exhibit any magnetic effect in the absence of an external magnetic field. The reason is that the atomic magnets are randomly oriented so the magnetic moment of the bulk of the substance remains zero.

Paramagnetism is temperature-dependent:

Curie’s Law: In 1895, Curie discovered experimentally that the magnetization I (magnetic moment per unit volume) of a paramagnetic substance is directly proportional to the magnetic intensity H of the magnetizing field and inversely proportional to the kelvin temperature T. That is \(I=C\left(\frac{H}{T}\right) \text {, }\)

where C is constant. This equation is known as Curie’s law and the constant C is called the Curie constant. The law, however, holds so long the ratio H/T does not become too large.

I cannot increase without limit. It approaches a maximum value corresponding to the complete alignment of all the atomic magnets constant in the substance.

Curie’s law can be expressed in an alternative form. We know that the magnetic susceptibility xm is defined as \(\chi_{\mathrm{m}}=\mathrm{I} / \mathrm{H}\)

Making this substitution in the above expression, we get \(\chi_{\mathrm{m}}=\mathrm{C} / \mathrm{T} \quad \Rightarrow \quad \chi_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}} .\)

Hysteresis: Retentivity and coercivity: Hysteresis curve: When a ferromagnetic substance is placed in a magnetic field, it is magnetized by induction. If we vary the magnetic intensity H of the magnetizing field, the intensity of magnetization and the flux density B in the (ferromagnetic) substance do not vary linearly with H.

In other words, the susceptibility Im (= I/H) and the permeability µ = (=B/H) of the substance are not constants, but vary with H and also depend upon the history of the substance.

The variation in I with variation in H is shown in the above figure. The point O represents the initial unmagnetized state of the substance (I = 0) and a zero magnetic intensity (H = 0). As H is increased, I increase (non-uniformly) along OA. At A the substance acquires a state of magnetic saturation. Any further increase in H does not produce any increase in I.

If now the magnetizing field H is decreased, the magnetization I of the substance also decreases following a new path AB (not the original path AO). Thus I lag behind H. When H becomes zero, I still have a value equal to OB. The magnetization remaining in the substance when the magnetizing field is reduced to zero is called “residual magnetism”.

The power of retaining this magnetism is called the “retentivity” or the remanence of the substance. Thus, the retentivity of a substance is a measure of the magnetization remaining in the substance when the magnetizing field is removed. In figure OB represents the retentivity of the substance If now the magnetizing field H is increased in the reverse direction, the magnetization I decrease along BC, still lagging behind H, until it becomes zero at C where H equals OC.

The value OC of the magnetizing field is called the “coercive “ or coercivity” of the substance. Thus, the coercivity of a substance is a measure of the reverse magnetizing field required to destroy the residual magnetism of the substance.

As H is increased beyond OC, the substance is increasingly magnetized in the opposite direction along CD, at D the substance is again magnetically saturated.

By taking H back from its maximum negative value (through zero) to its original maximum positive value, a symmetrical curve DEFA is obtained. At points B and E where the substance is magnetized in the absence of any external magnetizing field, it is said to be a “permanent magnet”.

It is thus found that the magnetization I (or of B) behind H is called “hysteresis”. The closed curve ABCDEFA which represents a cycle of magnetization of the substance is known as the “hysteresis curve (or loop)” of the substance. On repeating the process, the same closed curve is traced again but the portion OA is never obtained.

Hysteresis loss: A ferromagnetic substance consists of local regions called “domains”, each of which is spontaneously magnetized. In an unmagnetised substance the directions of magnetization in different domains are different so that, on average, the resultant magnetization is zero. It can be proved that the energy lost per unit volume of a substance in a complete cycle of magnetization is equal to the area of the hysteresis loop (I-Hcurve).

Difference in magnetic properties of soft iron and steel: A comparison of the magnetic properties of ferromagnetic substances can be made by the comparison of the shapes and sizes of their hysteresis loops.

In the figure are shown hystersis loops of soft iron and steel for the same values of I and H. We can draw the following conclusions regarding the magnetic properties of this substance from these loops.

The retentivity of soft iron (OB’) is greater than the retentivity of steel (OB).

The coercivity of soft iron (OC’) is less than the coercivity of steel (OC).

The hysteresis loss in soft iron is smaller than that in steel because the area of the soft iron is smaller than that of steel.

Curves between magnetic flux density B and magnetizing field H would reveal that the permeability of soft iron is greater than that of steel.

Section of magnetic materials:

The choice of a magnetic material for making a permanent magnet, electromagnet, core of transformer, or diaphragm of telephone earpiece can be decided from the hysteresis curve of the material.

Permanent magnets :

The material for a permanent magnet should have high retentivity so that the magnet is strong, and high coercivity so that the magnetization is not wiped out by stray external fields, mechanical ill-treatment, and temperature changes.

The hysteresis loss is immaterial because the material in this case is never put to cyclic changes of magnetization.

From these considerations, permanent magnets are made of steel. The fact that the retentivity of soft iron is a little greater than that of steel is outweighed by its much smaller coercivity, which makes it very easy to demagnetize.

Electromagnets: The material for the cores of electromagnets should have high permeability (or high susceptibility), especially at low magnetizing fields, and a high retentivity. Soft iron is a suitable material for electromagnets).

Transformer cores and telephone diaphragms: In these cases, the material goes through complete cycles of magnetization continuously. The material must therefore have a low hysteresis loss to have less dissipation of energy and hence a small heating of the material (otherwise the insulation of windings may break), a high permeability (to obtain a large flux density at low field), and a high specific resistance (to reduce eddy current loses).

Soft -iron is used for making transformer cores and telephone diaphragms: More effective alloys have now been developed for transformer cores. They are permalloys, mumetals, etc.

Electromagnet:

If we place a soft-iron rod in the solenoid, the magnetism of the solenoid increases hundreds of times. Then the solenoid is called an ‘electromagnet’. It is a temporary magnet.

An electromagnet is made by winding closely many turns of insulated copper wire over a soft-iron straight rod or a horse-shoe rod. On passing current through this solenoid, a magnetic field is produced in the space within the solenoid.

Example 27: A magnetizing field of 1600 Am–1 produces a magnetic flux of 2.4 × 10–5 wb in an iron bar of cross-sectional area 0.2cm2. Calculate the permeability and susceptibility of the bar.

Solution: \(\mathrm{B}=\frac{\Phi}{\mathrm{A}}=\frac{2.4 \times 10^{-5} \mathrm{~Wb}}{0.2 \times 10^{-4} \mathrm{~m}^2}=1.2 \mathrm{~Wb} / \mathrm{m}^2=1.2 \mathrm{~N} \mathrm{~A}^{-1} \mathrm{~m}^{-1} \text {. }\)

The magnetizing field (or magnetic intensity) H is 1600 Am – 1. Therefore, the magnetic permeability is given by

⇒ \(\mu=\frac{B}{H}=\frac{1.2 \mathrm{NA}^{-1} \mathrm{~m}^{-1}}{1600 \mathrm{Am}^{-1}}=7.5 \times 10^{-4} \mathrm{~N} / \mathrm{A}^2 \text {. }\)

Now, from the relation \(\mu=\mu_0\left(1+\chi_m\right)\) the susceptibility is given by \(\chi_m=\frac{\mu}{\mu_0}-1 .\) We known that µ 0 = 4π × 10 – 7 N/A2

therefore \(\chi_{\mathrm{m}}=\frac{7.5 \times 10^{-4}}{4 \times 3.14 \times 10^{-7}}-1=596 .\)

Example 28. The core of the toroid of 3000 turns has inner and outer radii of 11 cm and 12 cm respectively. A current of 0.6 A produces a magnetic field of 2.5 T in the core. Compute the relative permeability of the core. (µ 0 = 4π × 10–7 T m A – 1).

Solution: The magnetic field in the space enclosed by the windings of a toroid carrying a current I 0 is µ0 n i0 where n is the number of turns per unit length of the toroid and µ0 is the permeability of free space. If the space is filled by a core of some material of permeability µ, then the field is given by B = µ n i0 But µ = µ 0ur, where µ r is the relative permeability of the core material. Thus,

⇒ \(B=\mu_0 u_r n i_0 \quad \text { or } \quad \mu_r=\frac{B}{\mu_0 n i_0}\)

Here B= 2.5T, i0= 0.7 A and n \(\frac{3000}{2 \pi r} m^{-1} \text {, }\) where r is the mean radius of the toroid \(\left(\mathrm{r}=\frac{11+12}{2}=11.5 \mathrm{~cm} 11.5 \times 10^{-2} \mathrm{~m}\right)\)

Thus, \(\mu_r=\frac{2.5}{\left(4 \pi \times 10^{-7}\right) \times\left(3000 / 2 \pi \times 11.5 \times 10^{-2}\right) \times 0.7}=\frac{2.5 \times 11.5 \times 10^{-2}}{2 \times 10^{-7} \times 3000 \times 0.7}\)

µr = 684.5

Self Practice Problems

Question 39. A ferromagnetic material is heated above its curie temperature. Which one is a correct statement-

1. Ferromagnetic domains are perfectly arranged.
2. Ferromagnetic domains become random.
3. Ferromagnetic domains are not influenced.
4. Ferromagnetic material changes itself into diamagnetic material.

Answer: 2. Ferromagnetic domains become random.

Question 40. To protect a sensitive instrument from external magnetic jerks, it should be placed in a container made of-

1. Nonmagnetic substance
2. Diamagnetic substance
3. Paramagnetic substance
4. Ferromagnetic substance

Answer: 4. Ferromagnetic substance

Question 41. The ratio of the intensity of magnetization and magnetic field intensity is known as

1. Permeability
2. Magnetic flux
3. Magnetic susceptibility
4. Relative Permeability

Answer: 3. Magnetic susceptibility

Question 42. If a magnetic material, moves from stronger to weaker parts of a magnetic field, then it is known as

1. Diamagnetic
2. Paramagnetic
3. Ferromagnetic
4. Anti-ferromagnetic

Answer: 1. Diamagnetic

Question 43. The susceptibility of a magnetic substance is found to depend on temperature and the strength of the magnetizing field. The material is a-

1. Diamagnet
2. Ferromagnet
3. Paramagnet
4. Superconductor

Answer: 2. Ferromagnet

Question 44. Property possessed by ferromagnetic substance only is-

1. Attracting magnetic substance
2. Hysteresis
3. Susceptibility independent of temperature
4. Directional property

Answer: 2. Hysteresis

Solved Miscellaneous Problems

Problem 1. A bar magnet has a pole strength of 3.6 A-m and a magnetic length of 8 cm. Find the magnetic field at (a) a point on the axis at a distance of 6 cm from the center towards the north pole and (b) a point on the perpendicular bisector at the same distance.
Answer: 8.6 × 10–4 T; 7.7 × 10–5 T.

M = 3.6 × 8 × 102 A.m2

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{Mr}}{\left(\mathrm{r}^2-\mathrm{a}^2\right)^2}=8.6 \times 10^{-4} \mathrm{~T} \text {. }\)

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{M}{\left(r^2+a^2\right)^{3 / 2}}=7.7 \times 10^{-5} \mathrm{~T}\)

Problem 2. A loop in the shape of an equilateral triangle of side ‘a’ carries a current as shown in the figure. Find out the magnetic field at the center ‘C’ of the triangle.

Solution: B = B1 +B2 + B3 = 3B1

⇒ \(=3 \frac{\mu_0}{4 \pi} \times \frac{i}{\left(\frac{a}{2 \sqrt{3}}\right)} \times\left(\sin 60^{\circ}+\sin 60^{\circ}\right)=\frac{9 \mu_0 i}{2 \pi a}\)

Problem 3. Two long wires are kept along the x and y axes they carry currents  &  respectively in +ve x and B +ve y directions. Find at a point (0, 0, d).
Solution: \(\vec{B}=\vec{B}_1+\vec{B}_2=\frac{\mu_0}{2 \pi} \frac{\hat{i}}{d} \quad\left((-\hat{j})+\frac{\mu_0}{2 \pi} \frac{i}{d}(\hat{i})=\frac{\mu_0 I}{2 \pi d}(\hat{i}-\hat{j})\right.\)

Problem 4. Find ‘B’ at center ‘C’ in the following cases:

Answer:

1. \(\frac{\mu_0 I}{4 R} \otimes\)
2. \(\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}}\left(1+\frac{1}{\pi}\right) \otimes\)
3. \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\left(\frac{1}{2}+\frac{1}{\pi}\right) \otimes\)
4. \(\frac{\mu_0 I}{4}\left(\frac{1}{a}+\frac{1}{b}\right) \otimes\)
5. \(\frac{\mu_0 I}{4}\left(\frac{1}{a}-\frac{1}{b}\right) \otimes\)
6. \(\frac{\mu_0 I}{4}\left(\frac{1}{a}-\frac{1}{b}\right) \otimes\)
7. \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\left(1-\frac{1}{\pi}\right) \otimes\)
8. \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\left(1+\frac{1}{\pi}\right) \odot\)
9. \(\frac{\mu_0 I \theta}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right) \odot\)
10. Solution: \(B=\frac{\mu_0 I}{2 R} \times \frac{1}{2}=\frac{\mu_0 I}{4 R}\)
11. \(B=B_1+B_2=\left(\frac{\mu_0 I}{2 R} \times \frac{1}{2}\right)+\left(\frac{\mu_0}{4 \pi} \cdot \frac{I}{R}\right)=\frac{\mu_0 I}{4 R}\left(1+\frac{1}{\pi}\right)\)
12. \(B=B_1+B_2=\frac{\mu_0}{2 a} \times \frac{1}{2}+\frac{\mu_0}{2 b} \times \frac{1}{2}=\frac{\mu_0 I}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
13. \(B=B_1+B_2=\frac{\mu_0}{2 a} \times \frac{1}{2}+\frac{\mu_0}{2 b} \times \frac{1}{2}=\frac{\mu_0 I}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
14. \(B=B_1+B_2=\frac{\mu_0}{2 a} \times \frac{1}{2}+\frac{\mu_0}{2 b} \times \frac{1}{2}=\frac{\mu_0 I}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
15. \(B=B_1-B_2=\frac{\mu_0 I}{2 R}-\frac{\mu_0 I}{2 \pi R}=\frac{\mu_0 I}{2 R}\left(1-\frac{1}{\pi}\right)\)
16. \(B=B_1+B_2=\frac{\mu_0 I}{2 R}+\frac{\mu_0 I}{2 \pi R}=\frac{\mu_0 I}{2 R}\left(1+\frac{1}{\pi}\right)\)
17. \(B=B_1-B_2=\frac{\mu_0 I}{2 a}-\frac{\mu_0 \mathrm{I}}{2 b} \times \frac{\theta}{2 \pi}=\frac{\mu_0 \mathrm{I} \theta}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)\)

Problem 5. A thin solenoid of length 0.4 m and having 500 turns of wire carries a current 1A; then find the magnetic field on the axis inside the solenoid.
Answer: 5π × 10–4 T.

⇒\(B=\mu_0 n i=\frac{\mu_0 \mathrm{Ni}}{\ell}=5 \pi \times 10^{-4} \mathrm{~T} \text {. }\)

Problem 6. A charged particle of charge 2C is thrown vertically upwards with a velocity of 10 m/s. Find the magnetic force on this charge due to Earth’s magnetic field. Given vertical component of the earth = 3T and angle of dip = 37º.
Answer: 2 × 10 × 4 × 10–6 = 8 × 10–5 N towards west.

⇒ \(\tan 37^{\circ}=\frac{B_V}{B_H} \Rightarrow \quad B_H=\frac{4}{3} \times 3 \times 10^{-5} \mathrm{~T} \quad \Rightarrow \quad F=q v B_H=8 \times 10^{-5} \mathrm{~N}\)

Problem 7. A particle of charge q and mass m is projected in a uniform and constant magnetic field of B v strength B. The initial velocity vector makes an angle ‘θ’ with the. Find the distance traveled by the particle in time ‘t’.
Answer: vt

The speed of the particle does not change therefore distance covered by the particle is s = vt

Problem 8. Two long wires, carrying currents i 1 and i2, are placed perpendicular to each other in such a way that they just avoid contact. Find the magnetic force on a small length dl of the second wire situated at a distance l from the first wire.

Solution: The situation is shown in the figure. The magnetic field at the site of d, due to the first wire is, \(B=\frac{\mu_0 l_1}{2 \pi \ell}\)

This field is perpendicular to the plane of the figure going into it. The magnetic force on the
length dl is

⇒ \(\mathrm{dF}=\mathrm{i}_2 \mathrm{~d} \ell \mathrm{B} \sin 90^{\circ}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \mathrm{~d} \ell}{2 \pi \ell}\)

This force is parallel to the current i1.

Example 29: Curves in the graph shown to give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c, and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Which wire has the greatest radius?

1. a
2. b
3. c
4. d

Answer: Inside the cylinder

⇒ \(\begin{aligned}
& B .2 \pi r=\mu_0 \cdot \frac{I}{\pi R^2} \pi r^2 \\
& \Rightarrow \quad B=\frac{\mu_0 I}{2 \pi R^2} \cdot
\end{aligned}\)

⇒ \(B=\frac{\mu_0 I}{2 \pi r}\)

Inside cylinder Bα r and outside \(\mathrm{B} \propto \frac{1}{\mathrm{r}}\)

So at the surface nature of the magnetic field changes. Hence clear from the graph, wire ‘c’ has the greatest radius

2. Which wire has the greatest magnitude of the magnetic field on the surface?

1. a
2. b
3. c
4. d

Answer: The magnitude of the magnetic field is maximum at the surface of wire ‘a’.

3. The current density in wire a is

1. Greater than in wire c.
2. Less than in wire c.
3. Equal to that in wire c.
4. Not comparable to that of wire c due to lack of information.

Solution: Inside the wire.

⇒ \(B(r)=\frac{\mu_0}{2 \pi} \cdot \frac{I}{R^2} \cdot r=\frac{\mu_0 J r}{2}\)

⇒ \(\frac{d B}{d r}=\frac{\mu_0 \mathrm{~J}}{2}\)

i.e slope \(\propto \mathrm{J}\)

⇒ \(\propto\) current density

It can be seen that the slope of the curve for wire A is greater than for wire C.