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Difficulty: MediumElectromagnetic Induction

A flat circular coil of 8080 turns, each having an area of 0.02 m20.02\text{ m}^2, is placed perpendicularly in a uniform magnetic field. If the magnetic flux density decreases uniformly from 0.60 T0.60\text{ T} to 0 T0\text{ T} in 0.16 s0.16\text{ s}, what is the magnitude of the induced electromotive force (e.m.f.) in the coil?

  1. A
    0.075 V0.075\text{ V}
  2. 6.0 V6.0\text{ V}Answer
  3. C
    9.6 V9.6\text{ V}
  4. D
    15.0 V15.0\text{ V}

Answer

6.0 V6.0\text{ V}
According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is equal to the rate of change of magnetic flux linkage: E=NΔΦΔt=N⋅A⋅ΔBΔt\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N \cdot A \cdot \frac{\Delta B}{\Delta t}. Substituting N=80N = 80, A=0.02 m2A = 0.02\text{ m}^2, ΔB=0.60 T\Delta B = 0.60\text{ T}, and Δt=0.16 s\Delta t = 0.16\text{ s} yields E=80×0.02×0.600.16=6.0 V\mathcal{E} = 80 \times 0.02 \times \frac{0.60}{0.16} = 6.0\text{ V}.

Step-by-Step Solution

1
Calculate the change in magnetic flux density (ΔB\Delta B) and the change in magnetic flux per turn (ΔΦ\Delta \Phi).
ΔB=0.60 T−0 T=0.60 T\Delta B = 0.60\text{ T} - 0\text{ T} = 0.60\text{ T}, and ΔΦ=A×ΔB=0.02 m2×0.60 T=0.012 Wb\Delta \Phi = A \times \Delta B = 0.02\text{ m}^2 \times 0.60\text{ T} = 0.012\text{ Wb}.
Magnetic flux is defined as the product of the perpendicular magnetic flux density and the cross-sectional area.
2
Apply Faraday's Law of Electromagnetic Induction for an NN-turn coil: E=NΔΦΔt\mathcal{E} = N \frac{\Delta \Phi}{\Delta t}.
\mathcal{E} = 80 \times \frac{0.012\text{ Wb}}{0.16\text{ s}} = 80 \times 0.075\text{ V} = 6.0\text{ V}.
The induced e.m.f. is directly proportional to the total rate of change of magnetic flux linkage through all NN turns of the coil.

Key Concept

Faraday's Law of Electromagnetic Induction
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Topics

Electromagnetic Induction

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