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Q. 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 $0.1 \,T$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.7 \,s$ at a steady state. The magnitudes of current during interval is

Alternating Current

Solution:

$\phi_{0}=B A \cos \theta$
$=0.1 \times 10^{-2} \times \cos 45^{\circ}$
$\phi_{f}=0$
$\varepsilon=\left|\frac{\Delta \phi}{\Delta t}\right|=\frac{0.1 \times 10^{-2}}{0.7 \times \sqrt{2}}=1 mV $,
$ I=\frac{\varepsilon}{R}=\frac{1 mV }{0.5 \Omega}=2 mA$