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Difficulty: MediumElectromagnetic Induction

A coil consisting of 150150 turns is placed in a magnetic field. The magnetic flux passing through the coil decreases uniformly from 4.0×10−3 Wb4.0 \times 10^{-3}\text{ Wb} to 1.0×10−3 Wb1.0 \times 10^{-3}\text{ Wb} over a time interval of 0.015 s0.015\text{ s}. What is the magnitude of the average induced electromotive force in the coil in volts?

Answer: 30 V

Answer

The magnitude of the average induced electromotive force in the coil is 30 V30\text{ V}.
According to Faraday's law of electromagnetic induction, the magnitude of the average induced electromotive force in a coil with NN turns is given by E=NΔΦΔtE = N \frac{\Delta \Phi}{\Delta t}. Given N=150N = 150, ΔΦ=3.0×10−3 Wb\Delta \Phi = 3.0 \times 10^{-3}\text{ Wb}, and Δt=0.015 s\Delta t = 0.015\text{ s}, the magnitude of the induced e.m.f. is E=150×3.0×10−30.015=30 VE = 150 \times \frac{3.0 \times 10^{-3}}{0.015} = 30\text{ V}.

Step-by-Step Solution

1
Calculate the magnitude of the change in magnetic flux through the coil
ΔΦ=4.0×10−3 Wb−1.0×10−3 Wb=3.0×10−3 Wb\Delta \Phi = 4.0 \times 10^{-3}\text{ Wb} - 1.0 \times 10^{-3}\text{ Wb} = 3.0 \times 10^{-3}\text{ Wb}
Faraday's law relates induced e.m.f. directly to the rate of change of magnetic flux.
2
Apply Faraday's Law of Electromagnetic Induction equation for an N-turn coil
E=NΔΦΔtE = N \frac{\Delta \Phi}{\Delta t}
The total induced electromotive force in a coil is proportional to the number of turns and the rate of change of flux.
3
Substitute the given numerical values to compute the magnitude of the induced e.m.f.
E=150×3.0×10−3 Wb0.015 s=150×0.2 V=30 VE = 150 \times \frac{3.0 \times 10^{-3}\text{ Wb}}{0.015\text{ s}} = 150 \times 0.2\text{ V} = 30\text{ V}
Performing clean calculation without needing a calculator.

Key Concept

Faraday's Law of Electromagnetic Induction
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Topics

Electromagnetic Induction

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