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(A) Area of the square loop,$A = 10 \, cm 	imes 10 \, cm = 100 \, cm^2 = 10^{-2} \, m^2$.Initial magnetic flux linked with the loop,$\phi_1 = B_1 A \

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(A) Area of the square loop,$A = 10 \, cm 	imes 10 \, cm = 100 \, cm^2 = 10^{-2} \, m^2$.Initial magnetic flux linked with the loop,$\phi_1 = B_1 A \cos 	heta = 0.1 	imes 10^{-2} 	imes \cos 45^{\circ} = 0.1 	imes 10^{-2} 	imes \frac{1}{\sqrt{2}} = \frac{10^{-3}}{\sqrt{2}} \, Wb$.Final magnetic flux linked with the loop,$\phi_2 = 0 \, Wb$ (since $B_2 = 0$).The induced $EMF$ in the loop is $e = -\frac{d\phi}{dt} = -\frac{\phi_2 - \phi_1}{\Delta t} = -\frac{0 - \frac{10^{-3}}{\sqrt{2}}}{0.7} = \frac{10^{-3}}{0.7 	imes 1.414} \approx \frac{10^{-3}}{0.99} \approx 10^{-3} \, V$.The induced current in the loop is $I = \frac{e}{R} = \frac{10^{-3} \, V}{1 \, \Omega} = 10^{-3} \, A = 1.0 \, mA$.

$A$ square loop of wire,side length $10 \, cm$,is placed at an angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \, T$ to zero in $0.7 \, s$. The induced current in the loop (its resistance is $1 \, \Omega$) is ..... $mA$.

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