IB — physicsHard2 min read

Electromagnetic Induction and Faraday's Law

Faraday's law; Lenz's law; EMF; magnetic flux; generators and transformers

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# Electromagnetic Induction and Faraday's Law — IB Physics

1. Magnetic Flux

Φ=BAcos⁡θ\Phi = BA\cos\theta

Units: weber (Wb). Flux linkage = NΦN\Phi.

2. Faraday's Law

ε=−NΔΦΔt\boxed{\varepsilon = -N\frac{\Delta\Phi}{\Delta t}}

The induced EMF equals the negative rate of change of flux linkage.

To increase EMF: more turns, stronger field, faster change, larger area.

3. Lenz's Law

The induced current flows in a direction that opposes the change causing it. This ensures conservation of energy.

4. AC Generator

A coil rotating in a B field: Φ=BANcos⁡(ωt)\Phi = BAN\cos(\omega t) ε=BANωsin⁡(ωt)\varepsilon = BAN\omega\sin(\omega t)

Peak EMF: ε0=BANω\varepsilon_0 = BAN\omega. Produces sinusoidal AC.

5. Transformers

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

Ideal: VpIp=VsIsV_p I_p = V_s I_s (power conserved).

Step-up: Ns>NpN_s > N_p; Step-down: Ns<NpN_s < N_p.

Losses: eddy currents (reduced by lamination), resistance heating, flux leakage.

Worked Example: Example 1

Problem

A coil of 100 turns, area 0.04 m², in a field that changes from 0.5T to 0.1T in 0.1s. ε=N×ΔΦ/Δt=100×(0.5−0.1)(0.04)/0.1=100×0.016/0.1=16\varepsilon = N \times \Delta\Phi/\Delta t = 100 \times (0.5-0.1)(0.04)/0.1 = 100 \times 0.016/0.1 = 16 V

Solution

Worked Example: Transformer

Problem

Np=2000N_p = 2000, Ns=100N_s = 100, Vp=240V_p = 240 V. Vs=240×100/2000=12V_s = 240 \times 100/2000 = 12 V (step-down). If Is=5I_s = 5 A: Ip=VsIs/Vp=60/240=0.25I_p = V_s I_s/V_p = 60/240 = 0.25 A.

Solution

7. Practice Questions

    1. State Faraday's law and Lenz's law. (2 marks)
    1. A coil of 50 turns has flux through it changing at 0.02 Wb/s. Find the induced EMF. (1 mark)
    1. A step-up transformer has 200 primary turns and 5000 secondary turns. Input is 10V at 4A. Find the output voltage and current (ideal). (3 marks)

    Answers

    1. Faraday: induced EMF = rate of change of flux linkage. Lenz: induced current opposes the change that caused it.
    1. ε=N×dΦ/dt=50×0.02=1\varepsilon = N \times d\Phi/dt = 50 \times 0.02 = 1 V.

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Summary

  • Φ=BAcos⁡θ\Phi = BA\cos\theta; ε=−NΔΦ/Δt\varepsilon = -N\Delta\Phi/\Delta t
  • Lenz: opposes change (energy conservation)
  • Generator: ε=BANωsin⁡(ωt)\varepsilon = BAN\omega\sin(\omega t)
  • Transformer: Vs/Vp=Ns/NpV_s/V_p = N_s/N_p
Tags:
#ib#physics#electricity#electromagnetic-induction#faraday#lenz#transformer

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