Doubling the strength of the magnetic field through a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?

Question

BlackTom AI · Accepted Answer

Faraday’s law of electromagnetic induction gives the central relation: the induced emf ε in a loop equals the negative rate of change of magnetic flux Φ through the loop, ε = −N dΦ/dt, where N is the number of turns and Φ = B·A·cosθ (B = magnetic field magnitude, A = loop area, θ = angle between B and the loop normal).

1) Relevant concepts:
- Magnetic flux Φ = B A cosθ.  
- Induced emf magnitude |ε| = N |dΦ/dt|.  
- If A, N, θ, and the time interval of change are held constant, ε depends directly on the time-rate-of-change of B.

2) Step-by-step reasoning:
- With A, N, and θ constant, Φ(t) = A cosθ · B(t).  
- Differentiate: dΦ/dt = A cosθ · dB/dt.  
- Therefore |ε| = N A |cosθ| · |dB/dt|, showing |ε| ∝ |dB/dt|.

3) Simple calculation / intermediate steps (compare original and doubled cases):
- Suppose initially B changes by ΔB in time Δt, giving |ε1| = N A |cosθ| · (ΔB/Δt).  
- If the magnetic field strength is doubled over the same time (i.e., the change becomes 2ΔB in the same Δt), then |ε2| = N A |cosθ| · ((2ΔB)/Δt) = 2 · [N A |cosθ| (ΔB/Δt)] = 2|ε1|.

4) Why the correct answer is correct:
- Because induced emf is proportional to the rate of change of B, doubling the change in B (over the same time interval) doubles dB/dt and thus doubles the induced emf magnitude.

5) Why other options are incorrect:
- (b) Four times: would require ε ∝ B^2, which is false; ε is linear in B change.  
- (c) Half as much: contradicts the direct proportionality (would be for halving dB/dt).  
- (d) No change: incorrect because altering B changes the flux and thus the emf.

Conclusion: The induced emf becomes two times as much, so option a is correct.

Doubling the strength of the magnetic field through a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?单项选择题

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