Class 12 Physics
Electromagnetic Induction
Faraday’s laws of Induction
First law: -
- According to the first law an emf is induced in the circuit whenever the amount of magnetic flux linked with a circuit changes.
- Current was induced because of magnetic flux, as there is some current in the circuit therefore there will be some emf flowing in the circuit.
- Whenever the amount of magnetic flux linked with the circuit changes only at that time emf is induced.
- The induced emf will be there till there is change in the flux.
- When the magnet was moved then only there was change in the flux.
- As the magnet is moving the number of magnetic lines crossing the area is also changing.
- There is a change in the flux therefore there is induced emf.
- If the magnet is not moving, there will be no change in the amount of magnetic flux so there is no induced current.
Second law: -
- According to the second law the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
- Emf which is induced will depend upon rate at which the magnetic flux is changing.
- Mathematically:-
- Let Φ1 = flux at initial time t=0.
- Φ2 = flux after time t.
- Rate of change of flux=(Φ2 – Φ1)/t =dΦ/dt
- According to Faraday’s law:-
- Induced emf e ∝ (dΦ/dt)
- Experimentally the constant of proportionality was found to be 1 in all cases.
- Therefore e=(dΦ/dt)
- Consider a coil which has N number of turns;Therefore
- e = N(dΦ/dt)
Problem:- A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placedwith its plane perpendicular to the horizontal component of the earth’smagnetic field. It is rotated about its vertical diameter through 180°in 0.25 s. Estimate the magnitudes of the emf and current induced inthe coil. Horizontal component of the earth’s magnetic field at theplace is 3.0 × 10–5 T.
Answer:-
Initial flux through the coil,
ΦB (initial) = BA cos θ
= 3.0 × 10–5 × (π ×10–2) × cos 0º
= 3π × 10–7 Wb
Final flux after the rotation,
ΦB (final) = 3.0 × 10–5 × (π ×10–2) × cos 180°
= –3π × 10–7 Wb
Therefore, estimated value of the induced emf is,
ε = N (ΔΦ/Δt)
= 500 × (6π × 10–7)/0.25
= 3.8 × 10–3 V
I = ε/R = 1.9 × 10–3 A
Note that the magnitudes of ε and I are the estimated values. Theirinstantaneous values are different and depend upon the speed ofrotation at the particular instant.
Problem:- A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5s.The magnetic flux between the pole pieces is known to be 8x10-4 Wb. Estimate the emf induced in the wire?
Answer:-
dΦ = 8x10-4 Wb
dt =0.5s
e= - (dΦ/dt) =- (8x10-4) / (0.5)
=-1.6x10-3V
Class 12 Physics Electromagnetic Induction NCERT Chapter 6 Free Notes for Best Revision
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