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Question

# A square loop of side 10cm and resistance of 2Ω is moving in the region of uniform magnetic field of 0.1T with a velocity 2 m/s through a distance 1 m completely inside the field. The dissipated heat is?

A
Zero
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B
104J
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C
2×104J
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D
4×104J
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Solution

## The correct option is C 2×10−4JThe angle θ made by the area vector of the coil with the magnetic field 45°. Initial magnetic flux is Φ=BAcosθ FinalFluxΦmin=0 The change in flux is brought about in 0.70s. The magnitude of the induced emf is given by ε=|ΔΦB|ΔT=|ΔΦ−0|ΔT ε=10−3√2×0.7=1.0mV Magnitude of Current I=εR Therefore,Magnitude of Current I=1.0mV0.5Ω=2.0mA

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