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A square loop of side 10 cm and resistan...

A square loop of side `10 cm` and resistance `0.5 Omega` 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. The magnitude of current in this time-interval is.

Text Solution

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The initia magnetic flux is given by
`phi=BA cos theta`
Given , B=0.10 T, area of square loop =`10 xx 10 =100 cm ^2 = 10^(-2)m^2`
`therefore phi=(0.1xx10^(-2))/(sqrt2) Wb`
Final flux `phi_("min")=0`
The change in flux is brougth about in 0.70 s
the magnitude of the induced emf is
`e=(Deltphi)/(Deltat)=(|phi-0|)/(Deltat)=(10^(3-))/(sqrt2xx0.7)=1mV`
The magnitude of current is
`I=e/R=(10^(-3))/(0.5)=2mA`
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