Lenz's Law
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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.
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 = 
(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 .png)
(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.
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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. |
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- Find the current in the resistor
- Find the power dissipated in the resistor
- 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| = 
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)
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(c)Because of the induced current flowing in it, the rod experiences a force |
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= 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.
The mechanical power supplied by the external agent is
Pmech = 
...(ii)
Comparing equation (i) and (ii) we get
Pmech = Pelec
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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. |
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Solution
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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. |
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Solution
|E| = N
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
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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. |
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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 = 
or 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 .
