A square loop of side 0.10 m moves with a velocity of 2 m/s perpendicular to a uniform magnetic field of 0.05 T into a region where the field is present. What is the magnitude of the induced emf in the loop?
Answer
0.01 V
The correct answer is 0.01 V — Compute (varepsilon = B,l,v = 0.05\times0.10\times2 = 0.01;\text{V}.
Step-by-step solution
- Recognise that the emf induced in a moving conductor equals the product of magnetic field, length of side perpendicular to motion and velocity.
- Use the side length (l=0.10;\text{m}), velocity (v=2;\text{m/s}) and magnetic field (B=0.05;\text{T}).
- Compute (varepsilon = B,l,v = 0.05\times0.10\times2 = 0.01;\text{V}.
Common mistake: Using loop perimeter instead of side length
Concept tested: Induced emf in a moving conductor
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