A rectangular conducting loop of dimensions 8 cm × 3 cm has a small cut and moves out of a uniform magnetic field B = 0.3 T directed perpendicular to the loop. The loop speed is 2 cm·s⁻¹, normal to its shorter side. What emf develops across the cut? Image
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Question

A rectangular conducting loop of dimensions 8 cm × 3 cm has a small cut and moves out of a uniform magnetic field B = 0.3 T directed perpendicular to the loop. The loop speed is 2 cm·s⁻¹, normal to its shorter side. What emf develops across the cut? Image

Options

G1 Physics · Accepted Answer

Step 1: During exit, the flux through the loop decreases because the area inside the field changes. Step 2: Faraday’s law gives |ε| = B |dA/dt|. The changing area rate is dA/dt = ℓv, where ℓ = 3 cm = 0.03 m is the side perpendicular to motion. Step 3: v = 2 cm·s⁻¹ = 0.02 m·s⁻¹. Step 4: |ε| = Bℓv = 0.3 × 0.03 × 0.02 = 1.8 × 10⁻⁴ V. Final answer: 1.8 × 10⁻⁴ V (Option A).