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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?

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Solution

The correct option is **C** 2×10−4J

The 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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