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(A) The rate of change of magnetic field is $\frac{dB}{dt} = 20 \, mT/sec = 20 	imes 10^{-3} \, T/sec$.The area of the square loop is $A = (1 \, cm)^

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$1.25 	imes 10^{-7} \, A$,(anti-clockwise)

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(A) The rate of change of magnetic field is $\frac{dB}{dt} = 20 \, mT/sec = 20 	imes 10^{-3} \, T/sec$.The area of the square loop is $A = (1 \, cm)^2 = (10^{-2} \, m)^2 = 10^{-4} \, m^2$.The induced electromotive force $(EMF)$ in the loop is given by Faraday's law: $e = A \frac{dB}{dt} = 10^{-4} 	imes 20 	imes 10^{-3} = 2 	imes 10^{-6} \, V$.The total resistance of the square loop is $R_{total} = 4 	imes 4 \, \Omega = 16 \, \Omega$.The induced current in the loop is $i = \frac{e}{R_{total}} = \frac{2 	imes 10^{-6}}{16} = 0.125 	imes 10^{-6} \, A = 1.25 	imes 10^{-7} \, A$.According to Lenz's law,since the magnetic field directed into the page is increasing,the induced current will create a magnetic field directed out of the page to oppose this change. Therefore,the induced current flows in an anti-clockwise direction.

The magnetic field in the region shown in the figure increases at a constant rate of $20 \, mT/sec$. Each side of the square loop $ABCD$ has a length of $1 \, cm$ and a resistance of $4 \, \Omega$. Find the current in the wire $AB$ if the switch $S$ is closed.

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