Lenz's Law

Electromagnetic Induction of Class 12

LENZ'S LAW

The effect of the induced emf is such as to oppose the change in flux that produces it.

Lenz's Law

Fig. 4.5 (a)As the magnet approaches the loop, the positive flux through the loop increases. The induced currents sets up an induced magnetic field, Bind whose (negative) flux opposes this change. The direction of Bind is opposite to that of external field Bext due to the magnet.

(b)When the flux through the loop decreases as the magnet moves away from the loop, the flux due to the induced magnetic field tries to maintain the flux through the loop.

In order to incorporate Lenz's law in equation (4.1) the modern statement of Faraday's law of electromagnetic induction is

E = Lenz's Law(4.3)

The negative sign indicates that the induced emf opposes the change in magnetic flux that produces it.

If the single loop is replaced by a coil of N turns, then the net emf induced is given by

E = - N Lenz's Law(4.4)

Lenz's law is closely related to the law of conservation of energy and is actually a consequence of this general law of nature. As the north pole of the magnet moves toward the loop(see Fig. 4.5 a) a north pole appears on the upper surface of the loop which opposes the motion of the N-pole of the bar magnet. Thus, in order to move the magnet toward the loop with a constant velocity an external force is to be applied. The work done by this external force gets transformed into electric energy which produces induced current in the loop.

Example: 4.1

A metal rod of length slides at constant velocity v on conducting rails, placed in a uniform and constant magnetic field B perpendicular to the plane of the rails as shown in Fig. 4.6. A resistance R is connected between the two ends of the rail.

Lenz's Law

  1. Find the current in the resistor
  2. Find the power dissipated in the resistor
  3. Find the mechanical power needed to pull the rod.

Solution

In this case flux varies due to the change of area. Let x be the instantaneous distance moved by the rod, then the flux through the area enclosed is

ΦB = BA = Blx

The magnitude of the induced emf is

|E| = Lenz's Law

The flux is increasing because the area is increasing. The induced emf opposes the increase in flux, which means that the induced magnetic field is opposite to the external field. Thus, the induced current in the circuit is anticlockwise.

The magnitude of the current is

I = E/R = Bv/R

(b)The electric power dissipated in the resistor is

Pelec = I2 R = (Bv/R)2R..(i)

(c)Because of the induced current flowing in it, the rod experiences a force len's law = I due to the external field. The force  F is directed opposite to v. In order to move the rod at constant velocity an external force Fext = IlB must be applied to the right.

Lenz's Law

The mechanical power supplied by the external agent is

Pmech = Lenz's Law...(ii)

Comparing equation (i) and (ii) we get

Pmech = Pelec

Lenz's Law

Example 4.2

A square loop of side l moves at constant velocity v perpendicular to a uniform magnetic field as shown in the Fig. 4.7. Starting at the time at which it enters the field until it leaves the field, make plots of the variation in the flux through the loop and the emf induced in it as functions of time.

Lenz's Law

Solution

Lenz's Law

Example: 4.3

A circular coil of radius 5 cm consists of exactly 250 turns. A magnetic field is directed perpendicular to its plane (as shown in Fig. 4.8) is increasing at a rate of 0.6 T/s. If the resistance of the coil is 8Ω, find the current induced in the coil.

Lenz's Law

Solution

|E| = NLenz's Law

Here,N = 250; R = 5 cm = 5 × 10-2 m

dB/dt = +0.6  T/s

Therefore, E = (250)π (5 × 10-2)2 (0.6) = 1.18 V

The magnitude of current is

I = E/R = 1.18/8 = 0.147 A

Example: 4.4

A circular loop of area A and resistance R rotates with an angular velocity ω about an axis through its diameter as shown in Fig. 4.9. The plane of the loop is initially perpendicular to a constant magnetic field B. Find the induced current in the loop.

Lenz's Law

Solution

The instantaneous magnetic flux through the loop is

ΦB = BA cos θ

Since θ = ωt, therefore, ΦB = BA cos ωt

From Faraday's law, equation (4.3) is

E = Lenz's Lawor E = BAω sin ωt

The induced current is

I = E/R = (BAω/R)sin ωt

Both the induced emf and the current vary sinusoidally. The amplitude of the emf is BAω and that of the current is BAω/R .

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