Faraday's Law Physics Example 4

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Example 4

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A coil of 5050 turns has a magnetic flux that changes from 0.02 Wb0.02 \text{ Wb} to 0.08 Wb0.08 \text{ Wb} in 0.1 s0.1 \text{ s}. Find the induced EMF.

Solution

  1. 1
    Identify the change in magnetic flux: ΔΦ=0.080.02=0.06 Wb\Delta\Phi = 0.08 - 0.02 = 0.06 \text{ Wb}.
  2. 2
    Apply Faraday's law: E=NΔΦΔt=50×0.060.1\mathcal{E} = -N\frac{\Delta\Phi}{\Delta t} = -50 \times \frac{0.06}{0.1}.
  3. 3
    Calculate: E=50×0.6=30 V\mathcal{E} = -50 \times 0.6 = -30 \text{ V} The magnitude of the induced EMF is E=30 V|\mathcal{E}| = 30 \text{ V}.

Answer

E=30 V|\mathcal{E}| = 30 \text{ V}
Faraday's law states that the induced EMF equals the negative rate of change of magnetic flux times the number of turns. More turns and a faster flux change both produce a larger EMF. The negative sign reflects Lenz's law: the induced EMF opposes the change in flux.

About Faraday's Law

The induced EMF in a circuit equals the negative rate of change of magnetic flux through the circuit.

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