InterviewSolution
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InterviewSolution
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A square loop of side 10 cm and resistance 0.5Omega is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval. |
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Answer» Solution :The angle `theta` MADE by the area vector of the loop with the magnetic field is `45^(@)`. The initial magnetic flux is, `phi=BAcostheta=(0.1xx10^(-2))/(SQRT(2))Wb` Final flux, `phi_("min")=0` The change in flux is brought about in 0.70s. The magnitude of the induced emf is GIVEN by `epsilon=(|Deltaphi_(B)|)/(Deltat)=(|(phi-0)|)/(Deltat)=(10^(-3))/(sqrt(2)xx0.7)=1.0mV` And the magnitude of the current is `I=(epsilon)/(R)=(10^(-3)V)/(0.5Omega)=2mA` |
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