NEET Physics Class 12 Notes For Chapter 6 Electromagnetic Induction

Chapter 6 Electromagnetic Induction

1. Magnetic Flux

The concept of magnetic lines of force was first proposed by Faraday. Faraday tried to provide the lines of force a real form assuming them as stretched rubber bands.

• In modern physics, the concept of magnetic lines of force is used in visualization or explanation of principles only.
• The tangent drawn at any point on a line of force in a magnetic field shows the direction of the magnetic field at that point and the density of lines of force, i.e., the number of lines of force crossing normally a unit area indicates the intensity of the magnetic field.
• The lines of force in a uniform magnetic field are parallel straight lines equidistant from each other.
• Where the lines of force are near each other, B is higher and where the lines of force are far apart, B is lesser.
• The number of lines of force crossing a given surface is called flux from that surface.
• Suppose it is generally represented by Φ. Flux is a property of a vector field. If the vector field is a magnetic field, then the flux is called magnetic flux.

The magnetic flux crossing a certain area is equal to the scalar product of the vector field \((\vec{B})\) and the vector area \((\overrightarrow{\mathrm{dA}})\), that is

Magnetic flux \(\mathrm{d} \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dA}}=\mathrm{BdA} \cos \theta\)

where θ is the angle between the vector field \((\vec{B})\) and the vector area \(\overrightarrow{\mathrm{dA}}\).

Φ = ∫ \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dA}}\)

For a uniform magnetic field \((\vec{B})\) and plane surface \((\vec{A})\) = \((\vec{B})\).\((\vec{A})\) = BA cosθ

(Note: In real sense, area is a scalar quantity, but it can be treated as whose direction is in the direction of perpendicular pointing outward from the surface)

Magnetic flux is a scalar quantity.

If a plane surface of area A is imagined in a uniform magnetic field \((\vec{B})\), then

(1)when a surface is perpendicular to the magnetic field, the lines of force crossing that area, i.e., the magnetic flux is

Φ= BA because θ = 0, cos θ = 1

(2) If the surface is parallel to the field, then

θ = 90°, cos θ = 0

∴ Φ= BA cos 90 = 0

(3) when the normal to the surface makes an angle θ with the magnetic field, the magnetic flux is Φ = BA cosθ

Suppose the magnetic field is not uniform and the surface is in no plane. In that case, the element \(\overrightarrow{\mathrm{dA}}\) of the surface may be assumed as plane and magnetic field \((\vec{B})\) may also be assumed as uniform over his element.

Thus the magnetic flux coming out from this element is \(\mathrm{d} \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dA}}\)

Hence magnetic flux coming out from the entire surface \(\phi=\int_s \vec{B} \cdot \overrightarrow{d A}\)

For a closed surface the vector area element pointing outward is positive and the vector area element pointing inward is negative.

Magnetic lines of force are closed curves because free magnetic poles do not exist. Thus for a closed surface whatever the number of lines of force entering it, the same number of lines of force come out from it.

As a result for a closed curve \(\phi=\int_{\mathrm{s}} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dA}}=0 \text { or } \quad \nabla \cdot \vec{B}=0\)

Thus the net magnetic flux coming out of a closed surface is equal to zero.

For a normal plane surface in a magnetic field

Φ= BA

Hence B = \(\frac{\phi}{\mathrm{A}}\)

Thus the magnetic flux passing through normally from the surface of a unit is equal to magnetic induction B.

Therefore \(\frac{\phi}{\mathrm{A}}\) is also called flux density.

Unit Of Magnetic Flux – In the M.K.S. system the unit of magnetic flux is Weber (Wb) and in the C.G.S. system, the unit of magnetic flux is Maxwell.

1 weber = 10⁸ maxwell

The M.K.S unit of flux density or magnetic induction is Weber/m². It is also called tesla.

1 tesla = 1 weber/m²

The C.G.S unit of magnetic flux density is gauss.

1 gauss = 1 maxwell/cm⁸

1 tesla = 1 weber/m² = 10⁴ gauss

Dimensions Of Magnetic Flux: Φ = BA

∴ \([\phi]=\frac{N}{A-m} \times m^2=\frac{N-m}{A}=\frac{\left(kg-m-s^2\right) \times m}{A}\)

= \(k g-m^2-s^2-A-1 \quad=M^1 L^2 T-2 A-1\)

1. Magnetic Flux Solved Examples

Example 1. The plane of a coil of area 1m² and having 50 turns is perpendicular to a magnetic field of 3 x 10^-5 weber/m². The magnetic flux linked with it will be

1. 1.5 x 10^-3 weber
2. 3 x 10^-5 weber
3. 15 x 10^-5 weber
4. 150 weber

Solution:

Φ = NBA cosθ

but N = 50, B = 3 x 10^-5 wb/m²,

A = 1m², θ = 0 or Φ = NBA

= 50 × 3 ×10^-5× 1

= 150 × 10^-5 weber

∴ Answer will be (1)

Example 2. Consider the fig. A uniform magnetic field of 0.2 T is directed along the +x axis. Then what is the magnetic flux through the top surface of the figure?

1. Zero
2. 0.8 Wb
3. 1.0m Wb
4. -1.8m Wb

Solution:

The magnetic flux is Φ = BA cosθ

for the top surface, the angle between normal to the surface and the x-axis is θ = 60°, and B = 0.2 T, A = 10 x 10 x 10^-4 m²

Thus Φ = 0.2 x 10^-2 x cos (60)= 10^-3 Wb.

The correct answer is thus (3)

1.1 Faraday’s Laws Of Electromagnetic Induction

When magnetic flux passing through a loop changes with time or magnetic lines of force are cut by a conducting wire then an emf is produced in the loop or in that wire. This emf is called induced emf.

If the circuit is closed then the current will be called induced current.

magnetic flux = \(\int \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\)

The magnitude of induced emf is equal to the rate of change of flux w.r.t. time in the case of a loop.

In the case of a wire, it is equal to the rate at which magnetic lines of force are cut by a wire

E = \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

(–) the sign indicates that the emf will be induced in such a way that it will oppose the change of flux.SI unit of magnetic flux = Weber.

1.1 Faraday’s Laws Of Electromagnetic Induction Solved Examples

Example 1. A coil is placed in a constant magnetic field. The magnetic field is parallel to the plane of the coil as shown in the figure. Find the emf induced in the coil.

Solution:

Φ = 0 (always) since the area is perpendicular to the magnetic field.

∴ emf = 0

Example 2. Find the emf induced in the coil shown in the figure. The magnetic field is perpendicular to the plane of the coil and is constant.

Solution:

Φ = BA (always) = const.

∴ emf = 0

Example 3. Find the direction of the induced current in the coil shown in the figure. The magnetic field is perpendicular to the plane of the coil and it is increasing with time.

Solution:

Inward flux is increasing with time. To oppose it, an outward magnetic field should be induced.

Hence current will flow anticlockwise.

Example 4. Shows a coil placed in a decreasing magnetic field applied perpendicular to the plane of the coil. The magnetic field is decreasing at a rate of 10T/s. Find out the current in magnitude and direction

Solution: Φ = B.A

emf = A . \(\frac{\mathrm{d B}}{\mathrm {d t}}\) = 2 x 10 = 20 v

∴ i = 20/ 5 = 4 amp.

From Lenz’s law, the direction of the current will be anticlockwise.

Example 5. Figure shows a long current carrying wire and two rectangular loops moving with velocity v. Find the direction of current in each loop.

Solution:

In loop (1) no emf will be induced because there is no flux change.

In loop (2) emf will be induced because the coil is moving in a region of decreasing magnetic field inward in direction.

Therefore to oppose the flux decrease in inward direction, current will be induced such that its magnetic field will be inwards. For this direction the current should be clockwise.

2. Lenz’s Law (Conservation Of Energy Principle)

According to this law, emf will be induced in such a way that it will oppose the cause which has produced it. The figure shows a magnet approaching a ring with its north pole towards the ring.

We know that magnetic field lines come out of the north pole and magnetic field intensity decreases as we move away from the magnet. So the magnetic flux (here towards left) will increase with the approach of the magnet.

This is the cause of flux change. To oppose it, the induced magnetic field will be towards the right. For this, the current must be anticlockwise as seen by the magnet.

If we consider the approach of the North Pole to be the cause of flux change, Lenz’s law suggests that the side of the coil towards the magnet will behave as the North Pole and will repel the magnet.

We know that a current-carrying coil will behave like a North Pole if it flows anticlockwise. Thus as seen by the magnet, the current will be anticlockwise.

If we consider the approach of the magnet as the cause of the flux change, Lenz’s law suggests that a force opposite to the motion of the magnet will act on the magnet, whatever the mechanism.

Lenz’s law says that if the coil is set free, it will move away from the magnet because in doing so it will oppose the ‘approach’ of the magnet.

If the magnet is given some initial velocity towards the coil and is released, it will slow down. It can be explained as the following.

The current induced in the coil will produce heat. From energy conservation, if heat is produced there must be an equal decrease of energy in some other form, here it is the kinetic energy of the moving magnet.

Thus the magnet must slow down. So we can justify that Lenz’s law is the conservation of energy principle.

2.1 Induced Emf, Current And Change In A Circuit

If e.m.f induced in a circuit is E and the rate of change of magnetic flux is dΦ/dt, then from Faraday’s and Lenz’s law

E \(\propto-\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)\) or \(\mathrm{E}=-\mathrm{K}\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)\) where K is constant, equal to one.

Thus \(E=-\left(\frac{d \phi}{d t}\right)\)

If there are \(\mathrm{N}\) turns in the coil, then induced e.m.f will be

E = \(-N\left(\frac{d \phi}{d t}\right)\)

If the magnetic flux linked with the circuit changes from \(\phi_1 \quad to \quad \phi_2\), in time t, then induced e.m.f will be

E = \(-N\left(\frac{d \phi}{d t}\right)=-N\left(\frac{\phi_2-\phi_1}{\mathrm{t}}\right)\)

If the resistance of the circuit is R, then the current induced in the circuit will be

I = \(\frac{\mathrm{E}}{\mathrm{R}}=-\frac{\mathrm{N}\left(\phi_2-\phi_1\right)}{t R} \text { ampere } \quad=-\frac{\mathrm{N}}{\mathrm{R}}\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right) \text { ampere }\)

Induced current depends upon

1. The resistance of the circuit \(\mathrm{I} \propto \frac{1}{\mathrm{R}}\)
2. The rate of change of magnetic flux \(\mathrm{I} \propto\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)\)
3. The number of turns \((\mathrm{N}) ; \mathrm{I} \propto \mathrm{N}\)

If R = ∞, that is, the circuit is open, then the current will not flow and if the circuit is closed, then the current will flow in the circuit.

If change dq flows in the circuit in time dt, then the induced current will be

I = \(\left(\frac{d q}{d t}\right)\) or dq = I dt but \(I=\frac{1}{R}\left(\frac{d \phi}{d t}\right)\)

∴ dq = \(\frac{1}{R}\left(\frac{d \phi}{d t}\right) d t=\frac{1}{R} d \phi\) or \(q=\int \frac{d \phi}{R}=\frac{\phi_2-\phi_1}{R}\)

If N is the number of turns, then \(\mathrm{dq}=\frac{\mathrm{Nd} \phi}{R}, \mathrm{q}=\frac{\mathrm{N}\left(\phi_2-\phi_1\right)}{R}\)

The charge flowing due to induction does not depend upon the time but depends upon the total change in the magnetic flux. It does not depend upon the rate or time interval of the change in magnetic flux.

Whether the change in magnetic flux is rapid or slow, the charge induced in the circuit will remain the same.

Thus \(\mathrm{q} \propto \mathrm{d} \phi \text { or } \mathrm{q} \propto\left(\phi_2-\phi_1\right)\)

The induced charge depends upon the resistance of the circuit, i.e., q ∝ 1/R

If R = ∞ or the circuit is open, q = 0 that is charge will not flow in the circuit.

If R≠ ∞ or circuit is closed, then q ≠ 0, that is, the induced charge will flow in the circuit The e.m.f induced in the circuit does not depend upon the resistance of the circuit.

The e.m.f induced in the circuit depends upon the following factors –

1. Number of turns (N) in the coil,
2. Rate of change of magnetic flux,
3. Relative motion between the magnet and the coil,
4. The cross-sectional area of the coil,
5. The magnetic permeability of the magnetic substance or material placed inside the coil.

2.2 Fleming’s Right Hand Rule

This law is used for finding the direction of the induced e.m.f or current.

According to this law, if we stretch the right-hand thumb and two nearby fingers perpendicular to one another and the first finger points in the direction of the magnetic field and the thumb in the direction of motion of the conductor then the central finger will point in the direction of the induced current.

2.3 Direction Of Induced Emf And Current (Applications Of Lenz’s Law)

If the current flowing in a coil appears anti-clockwise, then that plane of the coil will behave like an N-pole.

If the current flowing in the coil appears clockwise, then that plane of the coil will behave like a S-pole.

If the north pole of a magnet is moved rapidly towards the coil, then according to Lenz’s law the induced current will flow in the coil in such a direction as to oppose the motion of the magnetic happen only when the face of the coil towards the magnet behaves as a north pole, that is, the induced current will appear flowing in an anti-clockwise direction as seen from the side of the magnet.

Thus a force of repulsion will be produced between the magnet and the coil coming near each other which will oppose the motion of the magnet.

Hence some mechanical work has to be done to move the magnet near the coil against this opposing force and this work (mechanical energy) is converted into current (electrical energy)

On bringing a south pole towards a coil the current induced in the coil will appear to flow in a clockwise direction as observed from the side of the magnet and the face of the coil towards the magnet will behave as a south pole.

On moving the north pole of a magnet away from the coil the current induced in the coil will appear to flow in the clockwise direction as seen from the side of the magnet and the face of the coil towards the magnet will behave as a south pole.

 

On moving the south pole of a magnet away from the coil the current induced in the coil will appear to flow in the anticlockwise direction as seen from the side of the magnet and the face of the coil towards the magnet will behave like a north pole.

If a magnet is allowed to drop freely through a copper coil, then an induced current will be produced in the coil. This current will oppose the motion of the magnet, as a result, the acceleration of the falling magnet due to gravity will be less than ‘g’.

If the coil is cut somewhere, then the emf will be induced in the coil only but current will not be induced. In the absence of induced current the coil will not oppose the motion of the magnet and the magnet will fall through the coil with the acceleration equal to g.

If a magnet is dropped freely in a hollow long metal cylinder, then the acceleration of the falling magnet will be less than gravitational acceleration.

As the magnet keeps on falling inside a tube, its acceleration will continue to decrease and after traversing a certain distance the acceleration will become zero. Now the magnet will fall with constant velocity. This constant velocity is called terminal velocity.

If a current-carrying coil is brought near another stationary coil, then the direction of induced current in the second coil will be in the direction of current in the moving coil.

If a current-carrying coil is taken away from a stationary coil, then the direction of induced current in the second coil will be opposite to the direction of current in the moving coil.

In the coils arranged in the following way, when the key K connected to the circuit of the primary coil, is pressed, an induced current is produced in the secondary coil.

The direction of the induced current in the secondary coil is opposite to the direction of the current in the primary coil. (From Lenz’s law) When the key is opened, then the current in the primary coil is reduced to zero but current is induced in the secondary coil.

The direction of this induced current is the same as the direction of the current in the primary coil. (Form Lenz’s law).

When current is passed through a coil, the current flowing through the coil changes. As a result, the magnetic flux linked with the coil changes. Due to this, a current is induced in the coil. If the current induced in the coil flows in the opposite direction of the applied current.

If the current flowing in the coil is decreased, then the current induced in the coil flows in the direction of the applied current so as to oppose the decrement of the applied current.

Two coils A and B are arranged as shown in the figure. On pressing the key K current flows through coil A in the clockwise direction and the current induced in coil B will flow in the anticlockwise direction. (From Lenz’s law)

On opening the key K the current flowing through coil A will go on decreasing. Thus the current induced in coil B will flow in the clockwise direction.

If current flows in a straight conductor from A to B as shown in the figure, then the direction of current induced in the loop placed near it will be clockwise. (From Lenz’s law).

Three identical circular coils A, B and C are arranged coaxially as shown in the figure. The coils A and C carry equal currents as shown. Coils B and C are fixed in position.

If coil A is moved towards B, then the current induced in coil B will be in a clockwise direction because the direction of the current induced in coil B will oppose the motion of coil A.

(The face of A towards B is the south pole, then the face of B towards A is the south pole). There is no relative motion between B and C so current will not be induced in coil B due to coil C.

3. Motional Emf

We can find emf induced in a moving rod by considering the number of lines cut by it per sec assuming there are ‘B’ lines per unit area. Thus when a rod of length l moves with velocity v in a magnetic field B, as shown, it will sweep area per unit time equal to Iv and hence it will cut B l v lines per unit time.

Hence emf induced between the ends of the rod = Bvl

Also emf= \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\).

Here Φ denotes flux passing through the area, swept by the rod. The rod sweeps an area equal to A/dt in time interval dt. Flux through this area = BA/dt.

Thus \(\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{B} \ell \mathrm{vdt}}{\mathrm{dt}}=\mathrm{Bv} \ell\)

If the rod is moving as shown in the following figure, it will sweep area per unit time = v l sinθ and hence it will cut B v l sinθ lines per unit time.

Thus emf = Bvl sinθ.

3.1 Explanation Of Emf Induced In Rod On The Basis Of Magnetic Force

If a rod is moving with velocity v in a magnetic field B,  as shown, the free electrons in a rod will experience a magnetic force in a downward direction and hence free electrons will accumulate at the lower end and there will be a deficiency of free electrons and hence a surplus of positive charge at the upper end.

These charges at the ends will produce an electric field in a downward direction which will exert an upward force on the electron.

If the rod has been moving for quite some time enough charges will accumulate at the ends so that the two forces qE and qvB will balance each other. Thus E = v B.

V_P – V_Q= V B l

The moving rod is equivalent to the following diagram, electrically.

Shows a closed coil ABCA moving in a uniform magnetic field B with a velocity v. The flux passing through the coil is a constant and therefore the induced emf is zero.

Now consider rod AB, which is a part of the coil. Emf induced in the rod =B L v

Suppose the emf induced in part ACB is E, as shown.

Since the emf in the coil is zero, Emf (in ACB) + Emf (in BA) = 0

or -E + vBL = 0 or E = vBL

Thus emf induced in any path joining A and B is the same, provided the magnetic field is uniform. Also, the equivalent emf between A and B is BLv (here the two emf’s are in parallel)

3.1 Explanation Of Emf Induced In Rod On The Basis Of Magnetic Force Solved Examples

Example 1. Find the emf induced in the rod in the following cases. The figures are self-explanatory.

Solution:

(1)here \(\overrightarrow{\mathrm{v}} \| \overrightarrow{\mathrm{B}} \text { so } \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}=0\)

emf = \(\vec{\ell} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0\)

(2)here \(\overrightarrow{\mathrm{v}} \| \vec{\ell}\)

so emf = \(\vec{\ell} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0\)

(3)here \(\overrightarrow{\mathrm{B}} \| \vec{\ell}\)

so emf = \(\vec{\ell} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0\)

Example 2. A circular coil of radius R is moving in a magnetic field B with a velocity v as shown in the figure.

Find the emf across the diametrically opposite points A and B.

Solution: emf = BVI_effective = 2 R v B

Example 3. An irregularly shaped wire AB moving with velocity v, as shown.

Find the emf induced in the wire.

Solution:

The same emf will be induced in the straight imaginary wire joining A and B, which is Bvl sin θ

Example 4. A rod of length l is kept parallel to a long wire carrying constant current i. It is moving away from the wire with a velocity v. Find the emf induced in the wire when its distance from the long wire is x.

Solution:

E = \(\mathrm{B} l \mathrm{~V}=\frac{\mu_0 \mathrm{i} l \mathrm{~V}}{2 \pi \mathrm{x}}\)

Or,

Emf is equal to the rate at which magnetic field lines are cut. In dt time the area swept by the rod is l v dt. the magnetic field lines cut in dt time = \(B l v d t=\frac{\mu_0 i l v d t}{2 \pi x} \text {. }\)

∴ The rate with which magnetic field lines are cut = \(\frac{\mu_0 \mathrm{i} / \mathrm{v}}{2 \pi \mathrm{x}}\)

Example 5. A rod of length l is placed perpendicular to a long wire carrying current i. The rod is moved parallel to the wire with a velocity v. Find the emf induced in the rod if its nearest end is at a distance ‘a’ from the wire.

Solution:

Consider a segment of rod of length dx, at a distance x from the wire. Emf induced in the segment

d \(\in=\frac{\mu_0 i}{2 \pi x} d x. v\)

∴ \(\epsilon=\int_a^{a+\ell} \frac{\mu_0 i v d x}{2 \pi x}=\frac{\mu_0 i v}{2 \pi} \ln \left(\frac{\ell+a}{a}\right)\)

Example 6. A rectangular loop is moving parallel to a long wire carrying current I with a velocity v. Find the emf induced in the loop if its nearest end is at a distance ‘a’ from the wire. Draw an equivalent electrical diagram.

Solution:

emf = 0.

e = \(\frac{\mu_0 i v}{2 \pi} \ell \mathrm{n}\left(\frac{a+b}{a}\right)\)

∴ \(V_Q-V_R=e, \quad V_P-V_s=e \quad \Rightarrow \quad i=\frac{e-e}{4 r}=0\)

4. Induced Emf Due To Rotation

4.1 Rotation Of The Rod

Consider a conducting rod of length l rotating in a uniform magnetic field.

Emf induced in a small segment of length dr, of the rod = v B dr = rω B dr

∴ emf induced in the rod = \(\omega \mathrm{B} \int_0^1 \mathrm{rdr}=\frac{1}{2} \mathrm{~B} \omega \mathrm{l}^2\)

the equivalent of this rod is as follows

or \(\varepsilon=\frac{\mathrm{d} \Phi}{\mathrm{dt}}\)

∴ \(\varepsilon=\frac{\mathrm{d} \Phi}{\mathrm{dt}}=\frac{\text { flux through the area swept by the rod in time } \mathrm{dt}}{\mathrm{dt}}=\frac{\mathrm{B} \frac{1}{2} \ell^2 \omega \mathrm{dt}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~B} \omega \ell^2\)

4.1 Rotation Of The Rod Solved Examples

Example 1. A rod PQ of length l is rotating about one end P in a uniform magnetic field B which is perpendicular to the plane of rotation of the rod. Point M is the midpoint of the rod. Find the induced emf between M and Q if that between P and Q = 100V.

Solution:

⇒ \(\mathrm{E}_{\mathrm{MO}}+\mathrm{E}_{\mathrm{PM}}=\mathrm{E}_{\mathrm{PQ}}\)

⇒ corner \(\rightarrow \frac{\mathrm{Bw} \ell^2}{2}=100\)

⇒ \(\mathrm{E}_{\mathrm{MO}}+\frac{\mathrm{B} \omega\left(\frac{\ell}{2}\right)^2}{2}=\frac{\mathrm{B} \omega \ell^2}{2}\)

⇒ \(\mathrm{E}_{\mathrm{MO}}=\frac{3}{8} \mathrm{~B} \omega \ell^2=\frac{3}{4} \times 100 \mathrm{~V}=75 \mathrm{~V}\)

Example 2. A rod of length L and resistance r rotates about one end as shown in the figure. Its other end touches a conducting ring of negligible resistance. A resistance R is connected between the centre and the periphery. Draw the electrical equivalence and find the current in the resistance R. There is a uniform magnetic field B directed as shown.

 

Solution:

currenti = \(\frac{\frac{1}{2} B \omega \ell^2}{R+r}\)

Example 3. In the above question find the force required to move the rod with constant velocity v, and also find the power delivered by the external agent.

Solution: The force needed to keep the velocity constant \(F_{\text {ext }}=i \ell B=\frac{B^2 \ell^2 v}{R+r}\)

Power due to external force \(=\frac{B^2 \ell^2 v^2}{R+r}=\frac{\varepsilon^2}{R+r}=i^2(R+r)\)

Note: that the power delivered by the external agent is converted into joule heating in the circuit. That means the magnetic field helps in converting the mechanical energy into joule heating.

Example 4. A rod PQ of mass m and resistance r is moving on two fixed, resistanceless, smooth conducting rails (closed on both sides by resistances R₁ and R₂). Find the current in the rod at the instant its velocity is v.

Solution:

i = \(\frac{B \ell V}{r+\frac{R_1 R_2}{R_1+R_2}}\)

this circuit is equivalent to the following diagram.

4.2. Emf Induced Due To Rotation Of A Coil Solved Examples

Example 1. A ring rotates with angular velocity ra about an axis perpendicular to the plane of the ring passing through the centre of the ring. A constant magnetic field B exists parallel to the axis. Find the emf induced in the ring

Solution:

Flux passing through the ring Φ = B. A is a constant here, therefore emf induced in the coil is zero. Every point of this ring is at the same potential, by symmetry.

4.3 Emf Induced In A Rotating Disc

Consider a disc of radius r rotating in a magnetic field B.

Consider an element dx at a distance x from the centre. This element is moving with speed v =ωx.

∴ Induced emf across dx = B(dx) v = Bdxωx = Bωxdx

∴ emf between the centre and the edge of the disc.

= \(\int_0^{\mathrm{r}} \mathrm{B} \omega \mathrm{xd} d x=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}\)

4.4. Rotation Of A Rectangular Coil In A Uniform Magnetic Field

If the figure is a conducting rectangular coil of area A and turns N is shown. It is rotated in a uniform magnetic field B about a horizontal axis perpendicular to the field with an angular velocity ω. The magnetic flux linked with the coil is continuously changing due to rotation.

θ is the angle between the perpendicular to the plane of the coil and the direction of the magnetic field.

The magnetic flux passing through the rectangular coil depends upon the orientation of the plane of the coil about its axis.

Magnetic flux passing through the coil \(\phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \quad \cos \theta=\mathrm{BA} \quad \cos \quad \omega \mathrm{t}\)

If there are N turns in the coil, then the flux linked with the coil Φ = BAN cosωt

Since Φ depends upon the time t, the rate of change of magnetic flux \(\frac{d \phi}{d t}=-B A N \omega \sin \omega t\)

According to Faraday’s law, the emf induced in the coil \(\epsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

or \(\epsilon=\text{BAN} \omega \sin \omega \mathrm{t}\)

BAN ω is the maximum value of emf induced,

Thus writing BANω  = ∈₀

∴ ∈ = ∈₀ sin ωt

This equation represents the instantaneous value of emf induced at time t.

If the total resistance of the circuit along with the coil is R, then the induced current due to alternating voltage

I = \(\frac{\epsilon}{R}=\frac{\epsilon_0}{R} \sin \omega t\)

or I = \(I_0 \sin \omega t\)

where \(I_0=\frac{\epsilon_0}{R}\) is the maximum value of current.

The magnetic flux linked with the coil and the emf induced at different positions of the coil in one rotational cycle is shown in the following table:

The variations of magnetic flux linked with the coil and induced e.m.f at different times given in the above table are shown in the following figure.

The phase difference between the instantaneous magnetic flux and induced emf is π/2.

The ratio of \(_\max\) and \(\phi_{\max }\) is equal to the angular velocity of the coil,

Thus \(\frac{\epsilon_{\max }}{\phi_{\max }}=\frac{\mathrm{NBA} \omega}{\mathrm{NBA}}=\omega\)

If \(\theta=\frac{\pi}{4}=45^{\circ}\), then \(\quad \Rightarrow \quad \phi=\frac{\mathrm{NBA}}{\sqrt{2}}\) and \(\epsilon=\frac{\mathrm{NBA} \omega}{\sqrt{2}}\)

In this case, the ratio of the induced emf and the magnetic flux is equal to the angular velocity of the coil.

Thus \(\frac{\epsilon}{\phi}=\frac{\mathrm{NBA}}{\sqrt{2}} / \frac{\mathrm{NBA}}{\sqrt{2}}=\omega\)

The direction of induced emf in the coil changes during one cycle so it is called alternating emf and current induced due to it is called alternating current. This is the principle of AC generator.

4.4. Rotation Of A Rectangular Coil In A Uniform Magnetic Field Solved Examples

Example 1. The phase difference between the emf induced in the coil rotating in a uniform magnetic field and the magnetic flux associated with it is

1. π
2. π/2
3. π/3
4. Zero

Solution:

Φ = NAB cosωt and ∈ = NAB ω sin ωt

Hence the phase difference between Φ and ∈ will be π/2.

∴ The answer will be (2)

Example 2. A coil has 20 turns and the area of each turn is 0.2 m². If the plane of the coil makes an angle of 60º with the direction of the magnetic field of 0.1 tesla, then the magnetic flux associated with the coil will be

1. 0.4 weber
2. 0.346 weber
3. 0.2 weber
4. 0.02 weber

Solution:

Φ = n(B da cosθ)

= 20 × 0.1 × 0.2 cos (90º– 60º) = 20 × 0.1 × 0.2 × \(\frac{\sqrt{3}}{2}\) = 0.346 weber

∴ The answer will be (2)

Example 3. A ring rotates with angular velocity ω about an axis in the plane of the ring and passes through the centre of the ring. A constant magnetic field B exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time.

Solution:

At any time t, Φ = BA cosθ = BA cosωt

Now induced emf in the loop = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{BA} \omega \sin \omega \mathrm{t}\)

If there are N turns emf = BAωN sin ωt

BA ωN is the amplitude of the emf e = e_m sin ωt

i = \(\frac{e}{R}=\frac{e_m}{R} \sin \omega t=i_m \sin \omega t\)

∴ \(i_m=\frac{e_m}{R}\)

The rotating coil thus produces a sinusoidally varying current or alternating current. This is also the principle used in the generator.

Example 4. Showing a wire frame PQSTXYZ placed in a time-varying magnetic field given as B=βt, where β is a positive constant. Resistance per unit length of the wire is λ. Find the current induced in the wire and draw its electrical equivalent diagram.

Solution:

Induced emf in part PQST = β a² (in an anticlockwise direction, from Lenz’s Law)

Similarly Induced emf in part TXYZ = β b² (in an anticlockwise direction, from Lenz’s Law)

The total resistance of the part PQST =λ4a.

The total resistance of the part PQST = l4b.

The equivalent circuit is shown in the following diagram.

writing KVL along the current flow \(\beta b^2-\beta a^2-\lambda 4 a i-\lambda 4 b i=0 \quad \Rightarrow \quad i=\frac{\beta}{4 \lambda}(b-a)\)

5. Fixed Loop In A Time-Varying Magnetic Field

Now consider a circular loop, at rest in a varying magnetic field. Suppose the magnetic field is directed inside the page and it is increasing in magnitude.

The emf induced in the loop will be \(\varepsilon=-\frac{d \phi}{d t}.\)

Flux through the coil will be \(\phi=-\pi r^2 B ; \frac{d \phi}{d t}=-\pi r^2 \frac{d B}{d t} ; \varepsilon=-\frac{d \phi}{d t}\)

∴ \(\varepsilon=\pi r^2 \frac{d B}{d t} . \)

∴ \(E 2 \pi r=\pi r^2 \frac{d B}{d t} \quad \text { or } \quad E=\frac{r}{2} \frac{d B}{d t}\)

Thus changing the magnetic field produces an electric field which is non-conservative in nature. Work done in the closed path on unit +ve charge is not zero. The lines of force associated with this electric field are closed curves.

6. Self Induction

Self-induction is the induction of emf in a coil due to its own current change. Total flux NΦ passing through a coil due to its own current is proportional to the current and is given as NΦ = L i where L is called the coefficient of self-induction or inductance.

The inductance L is purely a geometrical property i.e., we can tell the inductance value even if a coil is not connected in a circuit. Inductance depends on the shape and size of the loop and the number of turns it has.

If current in the coil changes by ΔI in a time interval At, the average emf induced in the coil is given as \(\varepsilon=-\frac{\Delta(\mathrm{N} \phi)}{\Delta \mathrm{t}}=-\frac{\Delta(\mathrm{LI})}{\Delta \mathrm{t}}=-\frac{\mathrm{L} \Delta \mathrm{I}}{\Delta \mathrm{t}}.\)

The instantaneous emf is given as \(\varepsilon=-\frac{\mathrm{d}(\mathrm{N} \phi)}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{LI})}{\mathrm{dt}}=-\frac{\mathrm{LdI}}{\mathrm{dt}}\)

S.I Unit of inductance is wb/amp or Henry(H)

L – self-inductance is +ve quantity.

L Depends On:

1. Geometry of loop
2. Medium in which it is kept. L does not depend upon the current.

L is a scalar quantity.

6.1 Self-Inductance Of Solenoid

Let the volume of the solenoid be V, and the number of turns per unit length be n.

Let a current I be flow in the solenoid. The magnetic field in the solenoid is given as B = \(\mu_0 n l\). The magnetic flux through one turn of solenoid \(\phi=\mu_0 n \text { I A. }\)

The total magnetic flux through the solenoid = \(N \phi=N \mu_0 n I A=\mu_0 n^2 \mid A I\)

∴ L = \(\mu_0 \mathrm{n}^2 I A=\mu_0 n^2 V\)

Φ = \(\mu_0 n i \pi r^2(n \ell)\)

L = \(\frac{\phi}{\mathrm{i}}=\mu_0 \mathrm{n}^2 \pi \mathrm{r}^2 \ell\).

Inductance per unit volume = \(\mu_0 \mathrm{n}^2\).

Self-inductance is the physical property of the loop due to which it opposes the change in current which means it tries to keep the current constant. The current can not change suddenly in the inductor.

7. Inductor

It is represented    by the electrical equivalence of a loop.

If current i through the inductor is increasing the induced emf will oppose the increase in current and hence will be opposite to the current.

If current i through the inductor is decreasing the induced emf will oppose the decrease in current and hence will be in the direction of the current.

Note: If there is a resistance in the inductor (resistance of the coil of the inductor) then:

Example: A B is a part of the circuit. Find the potential difference v_A – v_B if

1. current i = 2A and is constant
2. current i = 2A and is increasing at the rate of 1 amp/sec.
3. current i = 2A and is decreasing at the rate of 1 amp/sec

Solution:

L \(\frac{\mathrm{di}}{\mathrm{dt}}=1 \frac{\mathrm{di}}{\mathrm{dt}}\)

writing KVL from A to B \(V_A-1 \frac{d i}{d t}-5-2 i=V_B \text {. }\)

(1)Put \(i=2, \frac{d i}{d t}=0\); \(V_A-5-4=V_B\)

∴ \(V_A-V_B=9 \text { volt }\)

(2) Put \(\mathrm{i}=2, \frac{\mathrm{di}}{\mathrm{dt}}=1 ; \mathrm{V}_{\mathrm{A}}-1-5-4=\mathrm{V}_{\mathrm{B}}\) or \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=10\) Volt

(3) Put \(\mathrm{i}=2, \frac{\mathrm{di}}{\mathrm{dt}}=-1 ; \mathrm{V}_{\mathrm{A}}+1-5-2 \times 2=\mathrm{V}_{\mathrm{B}}\) or \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=8\) volt.

7.1 Energy Stored In An Inductor

If the current in an inductor at an instant is i and is increasing at the rate di/dt, the induced emf will oppose the current. Its behaviour is shown in the figure.

Power consumed by the inductor = i L \(\frac{di}{dt}\)

Energy consumed in dt time = i L \(\frac{di}{dt}\) dt

∴ total energy consumed as the current increases from 0 to I = \(\int_0^1 \mathrm{iLdi}=\frac{1}{2} \mathrm{LI}^2\)

= \(\frac{1}{2} \mathrm{Li}^2 \quad \Rightarrow \quad \mathrm{U}=\frac{1}{2} \mathrm{LI}^2\)

This energy is stored in the magnetic field with energy density \(\frac{d U}{d V}=\frac{B^2}{2 \mu}=\frac{B^2}{2 \mu_0 \mu_r} \quad \text { Total energy } U=\int \frac{B^2}{2 \mu_0 \mu_r} d V\)

Example 2. A circuit contains an ideal cell and an inductor with a switch. Initially, the switch is open. It is closed at t=0. Find the current as a function of time.

Solution:

ε = \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}} \quad \Rightarrow \int_0^{\mathrm{i}} \varepsilon \mathrm{dt}=\int_0^{\mathrm{i}} \mathrm{Ldi} \Rightarrow \quad \varepsilon \mathrm{t}=\mathrm{Li} \Rightarrow \mathrm{i}=\frac{\varepsilon \mathrm{t}}{\mathrm{L}}\)

Example 3. In the following circuit, the switch is closed at t = 0. Find the currents \(\mathrm{i}_1, \mathrm{i}_2, \mathrm{i}_3 \text { and } \frac{\mathrm{di}_3}{\mathrm{dt}} \text { at } \mathrm{t}=0\). Initially, all currents are zero.

Solution:

At t = 0

i₃ is zero since the current cannot suddenly change due to the inductor.

∴ i₁ = i₂(from KCL)

applyingKVL in the part ABEF we get \(i_1=\frac{\varepsilon}{2 R} i_2=, i_3=0, \frac{d i_3}{d t}=\frac{\epsilon}{2 L}\)

At t = ∞

i₃ will become constant and hence potential difference across the inductor will be zero. It is just like a simple wire and the circuit can be solved assuming it to be like shown in the following diagram.

⇒ \(\mathrm{i}_2=\mathrm{i}_3=\frac{\varepsilon}{3 \mathrm{R}}, \mathrm{i}_1=\frac{2 \varepsilon}{3 \mathrm{R}}, \frac{d \mathrm{i}_3}{\mathrm{dt}}=0\)

7.2 Growth Of Current In Series R–L Circuit

Figure shows a circuit consisting of a cell, an inductor L and a resistor R, connected in series.

Let the switch S be closed at t=0. Suppose at an instant current in the circuit is i which is increasing at the rate di/dt.

Writing KVL along the circuit, we have \(\varepsilon-L \frac{d i}{d t}-\mathrm{i} R=0\)

On solving we get, \(\mathrm{i}=\frac{\varepsilon}{\mathrm{R}}\left(1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right)\)

The quantity L/R is called the time constant of the circuit and is denoted by τ. The variation of current with time is as shown.

Note: 1. Final current in the circuit = \(\frac{\varepsilon}{\mathrm{R}}\), which is independent of L.

2. After one time constant, current in the circuit =63% of the final current (verify yourself)

3. More time constant in the circuit implies a slower rate of change of current.

4. If there is any change in the circuit containing the inductor then there is no instantaneous effect on the flux of the inductor.  L₁i₁ = L₂i₂

Example. At t = 0 switch is closed (shown in figure) after a long time suddenly the inductance of the inductor is made η times lesser \(\left(\frac{L}{\eta}\right)\) by pulling out the iron rod inserted in it then its initial value, find out instant current just after the operation.

Solution:

Using above result \(L_1 i_1=L_2 i_2 \quad \Rightarrow \quad \mathrm{i}_2=\frac{\eta \varepsilon}{R}\)

7.3 Decay Of Current In The Circuit Containing Resistor And Inductor

Let the initial current in the circuit be I₀. At any time t, let the current be i and let its rate of change at this instant be \(\frac{d i}{d t}\)

L\(\cdot \frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{iR}=0\)

⇒ \(\frac{\mathrm{di}}{\mathrm{dt}}=-\frac{\mathrm{iR}}{\mathrm{L}}\)

⇒ \(\int_{i_0}^1 \frac{d i}{i}=-\int_0^t \frac{R}{L} \cdot d t \quad \Rightarrow \quad \ln \left(\frac{i}{I_0}\right)=-\frac{R t}{L} \text { or } i=I_0 e^{\frac{-R t}{L}}\)

Current after one-time constant: i = I₀e^-1 = 0.37% of initial current.

Equivalent Self Inductance: \(\mathrm{L}=\frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}}{\mathrm{di} / \mathrm{dt}}\)…..(1)

Series Combination: \(V_A-L_1 \frac{d i}{d t}-L_2 \frac{d i}{d t}=V_B\) ….(2) from (1) and (2)

L = \(\mathrm{L}_1+\mathrm{L}_2\) (neglecting mutual inductance)

Parallel Combination:

From figure \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\mathrm{L}_1 \frac{\mathrm{di}_1}{\mathrm{dt}}=\mathrm{L}_2 \frac{\mathrm{di}_2}{\mathrm{dt}}\) also i = \(\mathrm{i}_1+\mathrm{i}_{\mathrm{z}}\)

or, \(\frac{d i}{d t}=\frac{d i_1}{d t}+\frac{d i_2}{d t}\) or \(\frac{V_A-V_B}{L}=\frac{V_A-V_B}{L_1}+\frac{V_A-V_B}{L_2}\)

∴ \(\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}\) (neglecting mutual inductance)

Example: An inductor having self-inductance L with its coil resistance R is connected across a battery of emfε. When the circuit is in a steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes n L (n> 1).

1. After insertion of the rod which of the following quantities will change with time?

1. The potential difference across terminals A and B.
2. Inductance.
3. The rate of heat produced in the coil

1. Only (1)
2. (1) and (3)
3. Only (3)
4. (1), (2) and (3)

Solution:

Inductance and potential differences across terminals will not change with time.

2. After insertion of the rod, current in the circuit:

1. Increases with time
2. Decreases with time
3. Remains constant with time
4. First decreases with time then becomes constant

Solution:

Even after the insertion of the rod, the current in the circuit will increase with time till a steady state is reached.

3. When again circuit is in a steady state, the current in it is:

1. \(K \varepsilon / R\)
2. \(\mid>E / R\)
3. \( I=\varepsilon / R\)
4. None of these

Solution:

At steady state inductor will offer zero resistance and hence I = \(\varepsilon / R\).

8. Mutual Inductance

Consider two arbitrary conducting loops 1 and 2. Suppose that I₁ is the instantaneous current flowing around loop 1. This current generates a magnetic field B₁ which links the second circuit, giving rise to a magnetic flux Φ₂ through that circuit.

If the current I₁ doubles, then the magnetic field B₁ doubles in strength at all points in space, so the magnetic flux Φ₂ through the second circuit also doubles.

Furthermore, it is obvious that the flux through the second circuit is zero whenever the current flowing around the first circuit is zero. It follows that the flux Φ₂ through the second circuit is directly proportional to the current I₁ flowing around the first circuit.

Hence, we can write Φ₂ =M₂₁ I₁ where the constant of proportionality M₂₁ is called the mutual inductance of circuit 2 with respect to circuit 1.

Similarly, the flux Φ₂ through the first circuit due to the instantaneous current I₂ flowing around the second circuit is directly proportional to that current, so we can write Φ₁ =M₁₂I₂ where M₁₂ is the mutual inductance of circuit 1 with respect to circuit 2.

It can be shown that M₂₁= M₁₂ (Reciprocity Theorem). Note that M is a purely geometric quantity, depending only on the size, number of turns, relative position, and relative orientation of the two circuits. The S.I. unit of mutual inductance is called Henry (H).

One Henry is equivalent to a volt-second per ampere:

Suppose that the current flowing around circuit 1 changes by an amount ΔI₁ in a small time interval Δt.

The flux linking circuit 2 changes by an amount ΔΦ₂=MΔI₁ in the same time interval.

According to Faraday’s law, an emf \(\varepsilon_2=-\frac{\Delta \phi_2}{\Delta \mathrm{t}}\) is generated around the second circuit due to the changing magnetic flux linking that circuit.

Since, \(\Delta \phi_2=M \Delta \mathrm{I}_1\), this emf can also be written \(\varepsilon_2=-M \frac{\Delta \mathrm{l}_1}{\Delta \mathrm{t}}\)

Thus, the emf generated around the second circuit due to the current flowing around the first circuit is directly proportional to the rate at which that current changes.

Likewise, if the current I₂ flowing around the second circuit changes by an amount ΔI₁ in a time interval Δt then the emf generated around the first circuit is \(\varepsilon_1=-M \frac{\Delta l_2}{\Delta t}\)

Note that there is no direct physical connection(coupling) between the two circuits: the coupling is due entirely to the magnetic field generated by the currents flowing around the circuits.

Note:

1. \(M \leq \sqrt{L_1 L_2}\)
2. For two coils in series if mutual inductance is considered then \(\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_1+\mathrm{L}_2 \pm 2 \mathrm{M}\)

Unit Of \(\mathbf{M}\): In M.K.S. system unit of mutual inductance is henry

M = \(\frac{E_B}{-\left(\mathrm{dI}_{\mathrm{A}} / \mathrm{dT}\right)}=\frac{\phi_B}{I_A}\)

∴ 1 henry = \(\frac{1 \text { volt }}{1 \text { ampere } / \mathrm{s}}=\frac{1 \text { weber }}{\text { ampere }}=\frac{\text { (joule } / \text { coulomb)s }}{\text { ampere }}=\mathrm{J} / \mathrm{A}^2\)

Dimensions Of M:

M = \(\frac{J}{A^2}=\frac{\text { joule }}{\text { ampere }^2}=\frac{\text { newton } \times \text { metre }}{\text { ampere }^2}=\frac{\mathrm{kg} \times \text { metre } \times \mathrm{sec}^{-2} \times \text { metre }}{\text { ampere }^2}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\)

Mutual inductance between the coils depends upon the number of turns in the coils, the area and the permeability of the core placed inside the coils. The larger the magnitude of M, the more is the emf induced in the secondary coil.

Out of the two coils coupled magnetically one coil can be taken as primary and the other coil as secondary. Thus mutual inductance MAB = MBA = M

Mutual inductance between two coaxial solenoids of length l and cross-sectional area A is M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{\ell}\) = where N₁ and N₂ are the number of turns in the two coils respectively.

If two coils are wound one over the other, then mutual inductance will be maximum and it will be less in other arrangements.

M and L have the following relation: \(M \propto \sqrt{L_1 L_2}\)

M = \(K \sqrt{L_1 L_2}\)

where K is a coupling constant of coils and its value varies from 0 to 1.

1. If K = 0, then there will be no coupling between the coils, that is magnetic flux produced by the primary coil is not linked with the secondary coil.
2. If K = 1, then both coils are coupled together with maximum transfer to energy, that is, the magnetic flux produced by the primary coil is totally linked with the secondary coil.

If two coils of self inductances L₁ and L₂ are coupled in series such that their windings are in the same sense and the mutual inductance between them is M, then the equivalent inductance will be L = L₁ + L₂ + 2M

If two coils are coupled in series such that their windings are in opposite sense then equivalent inductance will be L = L₁ + L₂ – 2M

8. Mutual Inductance Solved Examples

Example 1. A coil of radius 1 cm and 100 turns is placed at the centre of a long solenoid of radius 5 cm and 8 turns/cm. The value of the coefficient of mutual induction will be

1. 3.15 × 10^-5 H
2. 6 × 10^-5 H
3. 9 × 10^-5 H
4. Zero

Solution:

M = \(\mu_0 n_1 N_2 \pi \pi^2=4 \pi \times 10^{-7} \times 800 \times 100 \pi \times(0.01)^2=3.15 \times 10^{-5} \mathrm{H}\)

Hence the correct answer will be (1)

Example 2. The coefficients of self induction of two coils are 0.01 H and 0.03 H respectively. If they oppose each other then the resultant self-induction will be, if M = 0.01H

1. 2H
2. 0.02H
3. 0.02H
4. Zero

Solution:

L = L₁ + L₂ – 2M = 0.01 + 0.03 – 2 × 0.01

Hence the correct answer will be (3)

Example 3. Two insulated wires are wound on the same hollow cylinder, so as to form two solenoids sharing a common air-filled core. Let I be the length of the core, A the cross-sectional area of the core, N₁ the number of times the first wire is wound around the core, and N₂ the number of turns the second wire is wound around the core. Find the mutual inductance of the two solenoids,neglecting the end effects.

Solution:

If a current I₁ flows around the first wire then a uniform axial magnetic field of strength \(B_1=\frac{\mu_0 N_1 l_1}{\ell}\) is generated in the core. The magnetic field in the region outside the core is of negligible magnitude. The flux linking a single turn of the second wire is B₁A.

Thus, the flux linking all N₂ turns of the second wire is \(\phi_2=\mathrm{N}_2 \mathrm{~B}_1 \mathrm{~A}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{Al}_1}{\ell}=\mathrm{MI}_1\)

∴ M = \(\frac{\mu_0 N_1 N_2 \mathrm{~A}}{\ell}\)

As described previously, M is a geometric quantity depending on the dimensions of the core and the manner in which the two wires are wound around the core, but not on the actual currents flowing through the wires.

Example 4. Find the mutual inductance of two concentric coils of radii a₁ and a₂ (a₁<< a₂) if the planes of coils are the same.

Solution:

Let a current i flow in a coil of radius a₂

Magnetic field at the centre of coil = \(\frac{\mu_0 \mathrm{i}}{2 \mathrm{a}_2} \pi \mathrm{a}_1^2\)

or \(M i=\frac{\mu_0 i}{2 a_2} \pi a_1{ }^2\) or \(M=\frac{\mu_0 \pi a_1^2}{2 a_2}\)

Example 5. Solve the above question, if the planes of coil are perpendicular.

Solution:

Let a current i flow in the coil of radius a₁. The magnetic field at the centre of this coil will now be parallel to the plane of smaller coil and hence no flux will pass through it, M = 0.

Example 6. Solve the above problem if the planes of coils make θ angles with each other.

Solution:

If I current flows in the larger coil, magnetic field produced at the centre will be perpendicular to the plane of larger coil.

Now the area vector of smaller coil which is perpendicular to the plane of smaller coil will make an angle θ with the magnetic field.

Thus flux = \(\vec{B} \cdot \vec{A}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{a}_2} \cdot \pi \mathrm{a}_1^2 \cdot \cos \theta\) or \(\mathrm{M}=\frac{\mu_0 \pi \mathrm{a}_1^2 \cos \theta_1}{2 \mathrm{a}_2}\)

Example 7. The figure shows two concentric coplanar coils with radii a and b (a << b). A current i = 2t flows in the smaller loop. Neglecting self-inductance of the larger loop

1. Find the mutual inductance of the two coils
2. Find the emf induced in the larger coil
3. If the resistance of the larger loop is R find the current in it as a function of time

Solution:

(1) To find mutual inductance, it does not matter in which coil we consider current and in which flux is calculated (Reciprocity theorem) Let current I be flowing in the larger coil. Magnetic field at the centre = \(\frac{\mu_0 \mathrm{i}}{2 \mathrm{~b}}\)

flux through the smaller coil = \(\frac{\mu_0 i}{2 b} \pi a^2\)

∴ M = \(\frac{\mu_0}{2 b} \pi a^2\)

(2) |emf induced in larger coil| = M \(\mathrm{M}\left[\left(\frac{\mathrm{di}}{\mathrm{dt}}\right) \text { in smaller coil }\right](2)=\frac{\mu_0 \pi \mathrm{a}^2}{\mathrm{~b}}\)

(3) current in the larger coil = \(\frac{\mu_0 \pi \mathrm{a}^2}{\mathrm{bR}}\)

9. Eddy Current

When a conductor is placed in a changing magnetic field, induced EMF is produced in it. As a result local currents are produced in the conductor. These local currents are called eddy currents.

• If a conducting material is moved in a magnetic field, then eddy currents are also produced.
• Eddy currents flows in closed paths.
• There is a loss of energy due to eddy currents and it appears in the form of heat.
• In order to minimize the energy loss in the form of heat due to eddy currents the core of dynamo, motor or transformer is not taken as a single piece of soft iron but in the form of a peck of thin sheets insulated from each other by a layer of insulating varnish, called laminated core.
• This device increases the resistance for the eddy currents. In this way eddy currents are considerably reduced and loss of energy becomes less.

Uses Of Eddy Currents:

1. Moving coil galvanometer
2. Induction furnace
3. Dead beat galvanometer
4. Speedometer
5. Electric brakes

10. Generator Or Dynamo

A generator or dynamo is an electrical device which converts mechanical energy into electrical energy.

The working of generators is based on the principle of electromagnetic induction.

Generators Are Of Two Types:

1. A.C. Generator: If the current produced by the generator is alternating, then the generator is called A.C. generator.
2. D.C. Generator: If the current produced by the generator is direct current, then the generator is called a D.C. generator.

Generator consists of the following parts.

1. Armature (coil)
2. Magnet
3. Slip rings
4. Brushes

In D.C. generator commutator is used in place of slip rings.

In order to produce the magnetic field in big generators several magnetic poles are used. In these generators the armature coils are kept stationary and magnetic pole pieces are made to rotate around the armature.

The frequency of alternating current produced by generator of multi poles is = \(\frac{\text { number of poles } \times \text { rotational frequency }}{2}=\frac{\mathrm{Nn}}{2}\)

Energy Loss In Generators: The loss of energy is due to the following reasons

1. Flux leakage,
2. Copper losses,
3. Eddy’s current losses,
4. Hysteresis losses,
5. Mechanical losses

Efficiency Of Generator: Practical efficiency of a generator = \(\frac{\text { Electrical power generated by the generator }}{\text { Mechanical energy given to the generator }}\)

Practical efficiencies of big generators are about 92% to 95%.

11. Motor

It converts electrical energy into mechanical energy.

When a current carrying conductor (coil) is placed in a magnetic field, a couple acts on it which makes the coil to rotate.

Electric Motors Are Of Two Types:

1. Alternating current motor (AC motor)
2. Direct current motor (DC motor)

D.C. Motor Consists Of The Following Parts:

1. Armature
2. Magnet
3. Commentator
4. Brushes

Back E.M.F: When current from an external electric source is passed through the armature of the electric motor, the armature coil rotates in the magnetic field.

It cuts the magnetic lines of force as a result EMF is induced in it. According to Lenz’s law this induced emf opposes the rotation of the armature i.e., the emf induced works opposite to the emf applied by the external electric source and opposes the motion of the armature.

This induced emf is called back emf. The greater the speed of the armature coil, the greater is the back emf.

At the time of start of the motor back emf is almost zero and the current flowing in the motor is maximum. As the speed of the armature coil increases, back emf also increases.

When the coil increases, back emf also increases. When the coil attains maximum speed, the induced emf becomes constant and current reduced to minimum.

Back emf is directly proportional to the angular velocity ω of rotation of armature and the magnetic field B, i.e., for constant magnetic field back emf. e ∝ ω or e = Kω where K is a constant.

If E is applied emf, e is the back emf and R is the resistance of the coil (armature), then the current flowing through the coil will be

i = \(\frac{E-e}{R} \quad \text { or } \quad E=e+i R \quad \text { but } \quad e=K \omega\)

∴ \(i=\frac{E-K \omega}{R}\)

In the beginning, i.e. at the time of the start of the motor ω = 0

∴ i = \(\frac{E}{R}\)

In this case current will be maximum.

• As the armature coil is made from copper wire its resistance is very small. When the motor starts running, a very heavy current passes through the armature coil in the beginning.
• Due to which motor may get burnt.
• To prevent the motor from burning at the time of start a special variable resistance is connected in series with the armature, which is called starter.
• High resistance is connected in series with the armature coil with the help of starter at the time of start of the motor.
• As the motor starts picking up speed, the resistance is gradually reduced till it becomes zero.
• The starter is used in a high power motors but not in the low power motors because its coil starts rotating at a very high speed in a short time

Power of electric motor = ie

Efficiency of motor

η = \(\frac{\text { Work done by the motor }}{\text { Energy taken from the electric source by the motor }}=\frac{W}{P} \times 100 \%\)

or \(\eta =\frac{\text { Back emf }}{\text { Applied emf }} \times 100 \%=\frac{e}{E} \times 100 \%\)

Generally, the efficiency of the motor is from 80% to 90%.

11. Motor Solved Miscellaneous Problems

Problem 1. Find the emf across the points P and Q which are diametrically opposite points of a semicircular closed loop moving in a magnetic field as shown. Also, draw the electrical equivalent circuit of each branch.

Solution:

here \(\overrightarrow{\mathrm{v}} \| \vec{\ell}\)

So,emf = \(\vec{\ell} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=0\)

Induced emf = 0

Problem 2. Find the emf across the points P and Q which are diametrically opposite points of a semicircular closed loop moving in a magnetic field as shown. Also, draw the electrical equivalence of each branch.

Solution:

Induced emf = 2Bav

Problem 3. Figure shows a rectangular loop moving in a uniform magnetic field. Show the electrical equivalence of each branch.

Solution:

Problem 4. The figure shows a rod of length l and resistance r moving on two rails shorted by a resistance R. A uniform magnetic field B is present normal to the plane of rod and rails. Show the electrical equivalence of each branch.

Solution:

Problem 5. A rod PQ of length 2l is rotating about its midpoint C, in a uniform magnetic field B which is perpendicular to the plane of rotation of the rod. Find the induced emf between PQ and PC. Draw the circuit diagram of parts PC and CQ.

Solution:

∴ \(\mathrm{emf}_{\mathrm{PQ}}=0 ; \mathrm{emf}_{\mathrm{PC}}=\frac{\mathrm{B} \omega \ell^2}{2}\)

Problem 6. Which of the two curves shown has less time constant?

Solution:

curve 1

Problem 7. Find the mutual inductance of a straight long wire and a rectangular loop, as shown in the figure

Solution:

⇒ \(d \phi=\frac{\mu_0 i}{2 \pi r} \times b d r\)

⇒ \(\phi=\int_x^{x+a} \frac{\mu_0 i}{2 \pi r} \times b d r \quad \Rightarrow \quad M=\phi / l \quad \Rightarrow \quad M=\frac{\mu_0 b}{2 \pi} \ln \left(1+\frac{a}{x}\right)\)

Chapter 6 Electromagnetic Induction Key Concept

Magnetic flux is mathematically defined as \(\phi=\int \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\)

Faraday’s Laws Of Electromagnetic Induction

When magnetic flux passing through a loop changes with time or magnetic lines of force are cut by a conducting wire then an emf is produced in the loop or in that wire.

The magnitude of induced emf is equal to the rate of change of flux w.r.t. time in the case of a loop. In the case of a wire, it is equal to the rate at which magnetic lines of force are cut by a wire.

∴ \(\epsilon=-\frac{d \phi}{d t}\)

A negative sign implies that emf induces in such a way opposition to change in flux.

Lenz’s Law(based on the conservation of energy principle)

According to this law, emf will be induced in such a way that it will oppose the cause which has produced it.

If a wire AB is moving with constant velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\) then emf induced in any path joining A and B is same which is equal to \((\overrightarrow{\mathrm{AB}}) \cdot(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\) where L is the distance between A and B.

Induced Emf Due To Rotation Of Ring: Emf induced in a conducting rod of length l rotating with angular speed ra about its one end, in a uniform magnetic field B perpendicular to the plane of rotation is 1/2 B ωl².

EMF Induced In A Rotating Disc: Emf between the centre and the edge of the disc of radius r rotating in a magnetic field \(\mathrm{B}=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}\)

Self-Induction: Self-induction is the induction of emf in a coil due to its own current change. Total flux NΦ passing through a coil due to its own current is proportional to the current and is given as NΦ= L i where L is called the coefficient of self-induction or inductance.

The inductance L is purely a geometrical property.

If the current in the coil changes by ΔI in a time interval Δt, the average emf induced in the coil is given as \(\varepsilon=-\frac{\Delta(\mathrm{N} \phi)}{\Delta \mathrm{t}}=-\frac{\Delta(\mathrm{LI})}{\Delta \mathrm{t}}=-\frac{\mathrm{L} \Delta \mathrm{I}}{\Delta \mathrm{t}}\)

The instantaneous emf is gien as \(\varepsilon=-\frac{\mathrm{d}(\mathrm{N} \phi)}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{LI})}{\mathrm{dt}}=-\frac{\mathrm{Ldl}}{\mathrm{dt}}\)

Self inductance of solenoid = \(\mu_0 \mathrm{n}^2 \pi \mathrm{r}^2 \ell\)

Inductor: It is represented  by the electrical equivalence of the loop

Energy stored in an inductor = \(\frac{1}{2}\) LI²

Growth Of Current in Series R-L Circuit:

If a circuit consists of a cell, an inductor L a resistor R and a switch S, connected in series and the switch is closed at t = 0, the current in the circuit I will increase as i = \(\frac{\varepsilon}{R}\left(1-e^{\frac{-R t}{L}}\right)\)

1. Final current in the circuit = \(\frac{\varepsilon}{\mathrm{R}}\), which is independent of L.

2. After one time constant, the current in the circuit =63% of the final current.

Properties Of R-L Circuit

1. At t = 0, the inductor behaves like an open circuit.
2. At t = ∞ (after a very long time), the inductor behaves like a short-circuited wire.
3. Currently in inductor has not abruptly changed.

Decay Of Current In The Circuit Containing Resistor And Inductor:

Let the initial current in a circuit containing an inductor and resistor be I₀. Current at a time t is given as i = \(\mathrm{I}_0 e^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\)

Mutual Inductance: This is the induction of EMF in a coil (secondary) due to a change in current in another coil (primary). If the current in the primary coil is i, the total flux in the secondary is proportional to i, i.e. N Φ (in secondary) ∝i.

or N Φ (in secondary) = M i.

Equivalent Self Inductance: L = \(\frac{V_A-V_B}{d i / d t}\)…..(1)

Series Combination:

L = L₁ + L₂ (neglecting mutual inductance)

L = L₁ + L₂ + 2M (if coils are mutually coupled and they have winding in the same direction)

L = L₁ + L₂ – 2M (if coils are mutually coupled and they have to wind in opposite directions)

Parallel Combination: \(\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}\) (neglecting mutual inductance)

For two coils which are mutually coupled, it has been found that \(M \sqrt{L_1 L_2} \text { or } M=k \sqrt{L_1 L_2}\) where k is called coupling constant and its value is less than or equal to 1.

L.C. Oscillations: Let a capacitor be charged to Q and then connected in series with an inductor with the help of a switch as shown in the figure.

Let the switch be closed at t=0. Let at a time t=t, the charge on the capacitor be q and the current in the circuit be i where i = – \(\frac{dq}{dt}\).

Writing Kirchoff’s equation,we get \(\frac{q}{C}=L \frac{d i}{d t}=-L \frac{d^2 q}{d t^2} \text { or } \frac{d^2 q}{d t^2}+\frac{q}{L C}=0 \text {. }\)

Compared with the standard differential equation it can be easily seen that therefore time period T = \(2 \pi \sqrt{L C}\).

Example: Figure shows a conducting rod of negligible resistance that can slide on a smooth U-shaped rail made of wire of resistance 1 Ω/m. The position of the conducting rod at t = 0 is shown. A time t-dependent magnetic field B = 2t Tesla is switched on at t = 0.

1. The current in the loop at t = 0 due to induced emf is

1. 0.16 A, clockwise
2. 0.08 A, clockwise
3. 0.08 A, anticlockwise
4. zero

Solution:

⇒ \(\frac{\mathrm{dB}}{\mathrm{dt}}=2 \mathrm{~T} / \mathrm{s} \quad \mathrm{E}=-\frac{\mathrm{AdB}}{\mathrm{dt}}=-800 \times 10^{-4} \mathrm{~m}^2 \times 2=-0.16 \mathrm{~V}, \quad \mathrm{i}=\frac{0.16}{1 \Omega}\)

= \(0.16 \mathrm{~A}, \text { clockwise }\).

2. At t = 0, when the magnetic field is switched on, the conducting rod is moved to the left at a constant speed of 5 cm/s by some external means. The rod moves remaining perpendicular to the rails. At t = 2s, induced emf has magnitude.

1. 0.12 V
2. 0.08 V
3. 0.04 V
4. 0.02 V

Solution:

At t = 2 \(\mathrm{~s} \quad \mathrm{~B}=4 \mathrm{~T} ; \frac{\mathrm{dB}}{\mathrm{dt}}=2 \mathrm{~T} / \mathrm{s}\)

A = \(20 \times 30 \mathrm{~cm}^2\)

= \(600 \times 10^{-4} \mathrm{~m}^2 ; \frac{d A}{d t}=-(5 \times 20) \mathrm{cm}^2 / \mathrm{s} \quad=-100 \times 10^{-4} \mathrm{~m}^2 / \mathrm{s}\)

E = \(-\frac{d \varphi}{d t}=-\left[\frac{d(B A)}{d t}\right]=-\left[\frac{B d A}{d t}+\frac{A d B}{d t}\right]\)

= \(-\left[4 \times\left(-100 \times 10^{-4}\right)+600 \times 10^{-4} \times 2\right]=-[-0.04+0.120]=-0.08 \mathrm{~V}\)

Alternative: \(\phi=B A=2 t \times 0.2(0.4-v t)\)

= \(0.16 \mathrm{t}-0.4 \mathrm{vt}^2\)

E = \(-\frac{d \phi}{d t}=0.8 \mathrm{vt}-0.16\) at \(\mathrm{t}=2 \mathrm{~s}\)

E = -0.08 V

3. Following the situation of the previous question, the magnitude of the force required to move the conducting rod at a constant speed of 5 cm/s at the same instant t = 2s, is equal to

1. 0.16 N
2. 0.12 N
3. 0.08 N
4. 0.06 N

Solution: At t = 2s, length of the wire= (2 × 30 cm) + 20 cm = 0.8 m

Resistance of the wire = 0.8

Current through the rod = \(\frac{0.08}{0.8}=\frac{1}{10} \mathrm{~A}\)

Force on the wire = I l B = \(\frac{1}{10}\) × (0.2) × 4 = 0.08 N

The same force is applied on the rod in the opposite direction to make net force zero.

Problem: A rod PQ of length l is rotating about end P, with an angular velocity ω. Due to centrifugal forces, the free electrons in the rod move towards the end Q and an emf is created. Find the induced emf.

Solution:

The accumulation of free electrons will create an electric field which will finally balance the centrifugal forces and a steady state will be reached.

In the steady state m_eω²x = e E

∴ \(\left.V_p-V_Q=\int_{x=0}^{x=\ell} \bar{E} \cdot d \bar{x}=\int_0^{\ell} \frac{m_e \omega^2 x}{e} d x=\frac{m_e \omega^2 \ell^2}{2 e} \approx 10^{-12} V \quad \text { (If } \omega=1, \ell=1\right)\)

This value is very small.

Chapter 6 Electromagnetic Induction Questions And Answers

Question 1. A thin semicircular conducting ring of radius R is falling with its plane vertically in a horizontal magnetic induction B. At the position MNQ the speed of the ring is v then the potential difference developed across the ring is:

1. Zero
2. \(\frac{B v \pi R^2}{2}\)
3. π RBV and Q have a higher potential
4. 2 RBV and Q are at higher potential.

Answer: 4. 2 RBV and Q are at higher potential.

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i.e., e_MNQ = e_MQ = Bvl = Bv (2R)

[l = MQ = 2R]

Therefore, the potential difference developed across the ring is 2RBv with Q at a higher potential.

Question 2. A cylindrical space of radius R is filled with a uniform magnetic induction B parallel to the axis of the cylinder. If B changes at a constant rate, the graph showing the variation of the induced electric field with distance r from the axis of the cylinder is

Answer: 1

EMF = \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{dB} \pi \mathrm{r}^2}{\mathrm{dt}}=-\pi \mathrm{r}^2 \frac{\mathrm{dB}}{\mathrm{dt}} \text { or } \mathrm{E}=\left(\frac{\mathrm{EMF}}{2 \pi \mathrm{r}}\right)=\left(\frac{\mathrm{dB}}{\mathrm{dt}}\right) \frac{\mathrm{r}}{2}
\)

or \(E \propto r\) for \(r \leq R\).

E \(\propto \frac{1}{r}\) for r>R.

Question 3. In a cylindrical region uniform magnetic field which is perpendicular to the plane of the figure is increasing with time and a conducting rod PQ is placed in the region. If C is the centre of the circle then

1. P will be at a higher potential than Q.
2. Q will be at a higher potential than P.
3. Both P and Q will be equipotential.
4. No EMF will be developed across the rod as it is not crossing/cutting any line of force.

Answer: 2. Q will be at a higher potential than P.

If the circuit Q C P containing rod PQ is completed then the direction of induced current will be from Q to C to P hence Q will be at a higher potential than P.

Question 4. In a series L-R growth circuit, if the maximum current and maximum induced emf in an inductor of inductance 3mH are 2A and 6V respectively, then the time constant of the circuit is:

1. 1 ms.
2. 1/3 ms.
3. 1/6 ms
4. 1/2 ms

Answer: 1. 1 ms.

R = \(\frac{V}{I}\)

∴ \(\tau=\frac{L}{R}=1 \mathrm{~ms}\)

Question 5. Two coils of self-inductance 100 mH and 400 mH are placed very close to each other. Find the maximum mutual inductance between the two when 4 A current passes through them

1. 200 mH
2. 300 mH
3. 100√2 mH
4. None of these

Answer: 1. 200 mH

∴ \(M_{\max }=\sqrt{L_1 L_2}=\sqrt{100 \times 400} \mathrm{mH}=200 \mathrm{mH} \text {. }\)

Question 6. A nonconducting ring of radius R and mass m having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time-varying uniform magnetic field B = 4t² is switched on at time t=0. The coefficient of friction between the ring and the table, if the ring starts rotating at t =2 sec, is :

1. \(\frac{4 q m R}{g}\)
2. \(\frac{2 q m R}{g}\)
3. \(\frac{8 q R}{m g}\)
4. \(\frac{q R}{2 m g}\)

Answer: 3. \(\frac{8 q R}{m g}\)

\(\int E d \ell=-\frac{d \phi}{d t}\)

E \(\times 2 \pi R=\pi R^2 \frac{d B}{d t}\)

E = \(\frac{R}{2} \times 8 t=R 8\)

(qE)R  = \((\mu \mathrm{mg}) R\)

μ \(=\frac{8 q R}{m g}\)

Question 7. In the figure shown a square loop PQRS of side ‘a’ and resistance ‘r’ is placed near an infinitely long wire carrying a constant current I. The sides PQ and RS are parallel to the wire. The wire and the loop are in the same plane. The loop is rotated by 180° about an axis parallel to the long wire and passes through the midpoints of the side QR and PS. The total amount of charge which passes through any point of the loop during rotation is:

1. \(\frac{\mu_0 \mathrm{Ia}}{2 \pi \mathrm{r}} \ell \mathrm{n} 2\)
2. \(\frac{\mu_0 \text { Ia }}{\pi}\) en2
3. \(\frac{\mu_0 I a^2}{2 \pi r}\)
4. Cannot be found because the time of rotation is not given.

Answer: 2. \(\frac{\mu_0 \text { Ia }}{\pi}\) en2

q = \(\int \mathrm{Idt}=\int-\frac{1}{\mathrm{r}} \frac{\mathrm{d} \phi}{\mathrm{dt}} \mathrm{dt}=-\frac{\Delta \phi}{\mathrm{r}}=\frac{\mu_0 \mathrm{Ia}}{\pi \mathrm{r}} \ell \mathrm{n} 2\).

Question 8. Shows a conducting loop being pulled out of a magnetic field with a constant speed v. Which of the four plots shown in fig. may represent the power delivered by the pulling agent as a function of the constant speed v?

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

Answer: 2. B

P = \(\mathrm{F} . \mathrm{V}=\mathrm{Bi} \ell \mathrm{V}=\mathrm{B}\left(\frac{\mathrm{Bv} \ell}{\mathrm{R}}\right) \ell \mathrm{V}, \mathrm{P} \propto \mathrm{V}^2\)

Question 9. A uniform magnetic field, B = B₀ t (where B₀ is a positive constant), fills a cylindrical volume of radius R, then the potential difference in the conducting rod PQ due to the electrostatic field is:

1. \(\mathrm{B}_0 \ell \sqrt{\mathrm{R}^2+\ell^2}\)
2. \(\mathrm{B}_0 \ell \sqrt{\mathrm{R}^2-\frac{\ell^2}{4}}\)
3. \(\mathrm{B}_0 \ell \sqrt{\mathrm{R}^2-\ell^2}\)
4. \(B_0 R \sqrt{R^2-\ell^2}\)

Answer: 3. \(\mathrm{B}_0 \ell \sqrt{\mathrm{R}^2-\ell^2}\)

∫\(\mathrm{Ed} \ell=\varepsilon, \mathrm{E}=\frac{\mathrm{r}}{2} \frac{\mathrm{dB}}{\mathrm{dt}}, \quad \mathrm{E} \cos \theta=\frac{\mathrm{r} \cos \theta}{2} \quad \mathrm{~B}_0=\frac{\mathrm{h}}{2} \quad \mathrm{~B}_0\)

∴ \(V_Q-V_P=\left(\frac{\mathrm{h}}{2} \mathrm{~B}_0\right) 2 \ell=\mathrm{B}_0 \ell \sqrt{\mathrm{R}^2-\ell^2}\)

Question 10. An LR circuit has L = 1 H and R = 1 Ω. It is connected across an emf of 2 V. The maximum rate at which energy is stored in the magnetic field is:

1. 1 W
2. 2 W
3. 1/4 W
4. 4 W

Answer: 1. 1 W

U = \(\frac{1}{2} \mathrm{LI}^2\)

⇒ \(\frac{\mathrm{dU}}{\mathrm{dt}}=\mathrm{LI} \frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{RI}_0^2\left(1-\mathrm{e}^{-\mathrm{V}^2}\right) \mathrm{e}^{-\nu / t}\)

⇒ \(\frac{d U}{d t}\) is maximum when \(e^{-s t}=\frac{1}{2} \text { or }\left(\frac{d U}{d t}\right)_{\max }=\frac{E^2}{4 \mathrm{R}}=1 \mathrm{~W} \text {. }\)

Question 11. When induced emf in the inductor coil is 50% of its maximum value then stored energy in the inductor coil in the given circuit will be:

1. 2.5 mJ
2. 5mJ
3. 15 mJ
4. 20 mJ

Answer: 1. 2.5 mJ

E = \(\frac{1}{2} \mathrm{LI}^2 \quad E=\frac{1}{2} L \frac{\mathrm{V}^2}{\mathrm{R}^2}\)

= \(\frac{1}{2} \times 5 \times 10^{-3} \times(1)^2\)

= 2.5 mJ.

Question 12. A bar magnet is released from rest coaxially along the axis of a very long, vertical copper tube. After some time the magnet

1. Will move with an acceleration g
2. Will move with almost constant speed
3. Will stop in the tube
4. Will oscillate

Answer: 2. Will move with almost constant speed

Since the tube is very long the force on the magnet due to the induced current will continue to oppose its motion till it acquires a constant speed.

Question 13. As shown in the figure P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed a clockwise current l_P flows in P (as seen by E) and an induced current I_Q1 flows in Q. The switch remains closed for a long time. When S is opened, a current I_Q2 flows in Q. Then the directions of I_Q1 and I_Q2 (as seen by E) are

1. Respectively clockwise and anti-clockwise
2. Both clockwise
3. Both anti-clockwise
4. Respectively anti-clockwise and clockwise.

Answer: 4. Respectively anti-clockwise and clockwise.

When switch S is closed magnetic field lines passing through Q increases in the direction from right to left. So, according to Lenz’s law induced current in Q i.e. I_Q1 will flow in such a direction that the magnetic field lines due to I_Q1 pass from left to right through Q.

This is possible when I_Q1 flows in an anticlockwise direction as seen by E. The Opposite is the case when switch S is opened i.e. I_Q2 will be clockwise as seen by E.

Question 14. An infinitely long cylindrical conducting rod is kept along the +Z direction. A constant magnetic field is also present in the +Z direction. The current induced will be

1. 0
2. Along +z direction
3. Along clockwise as seen from +Z
4. Along anticlockwise as seen from +Z

Answer: 1. 0

Zero, as there is no flux change.

Question 15. Which of the field patterns given below is valid for electric fields as well as for magnetic fields?

Answer: 3

True for induced electric field and magnetic field.