Class 12 Physics Electromagnetic Induction | Faradays laws of 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 Φ
_{1}= flux at initial time t=0. - Φ
_{2}= flux after time t. - Rate of change of flux=(Φ
_{2}– Φ_{1})/t =dΦ/dt

- Let Φ
- 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

__Answer:-__

Initial flux through the coil,

Φ_{B} (initial) = BA cos θ

= 3.0 × 10^{–5} × (π ×10^{–2}) × cos 0º

= 3π × 10^{–7} Wb

Final flux after the rotation,

Φ_{B} (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

__Answer:-__

dΦ = 8x10^{-4} Wb

dt =0.5s

e= - (dΦ/dt) =- (8x10^{-4}) / (0.5)

=-1.6x10^{-3}V

.