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(B) According to Faraday's law of induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$,where $\phi = B \cdo

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$I_1 > I_2$

Vedclass · Answer

(B) According to Faraday's law of induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$,where $\phi = B \cdot A$ is the magnetic flux.Since the magnetic field $B$ is uniform and decreasing at a constant rate,the rate of change of flux is $\frac{d\phi}{dt} = A \frac{dB}{dt}$.The induced current is $I = \frac{\varepsilon}{R} = \frac{A}{R} \left| \frac{dB}{dt} \right|$,where $R$ is the resistance of the wire frame.In case $I$,the total area enclosed by the loop is $A_1 = L^2 + l^2$. The total length of the wire is $P_1 = 4L + 4l$,so the resistance is $R_1 = \rho \frac{P_1}{a} = \rho \frac{4(L+l)}{a}$ (where $\rho$ is resistivity and $a$ is cross-sectional area).In case $II$,the total area enclosed by the loop is $A_2 = L^2 - l^2$. The total length of the wire is $P_2 = 4L + 4l$,so the resistance is $R_2 = \rho \frac{4(L+l)}{a} = R_1$.Since $A_1 > A_2$ and $R_1 = R_2$,it follows that $I_1 = \frac{A_1}{R_1} |\frac{dB}{dt}| > I_2 = \frac{A_2}{R_2} |\frac{dB}{dt}|$.Therefore,$I_1 > I_2$.

The adjoining figure shows two different arrangements in which two square wire frames are placed in a uniform constantly decreasing magnetic field $B.$ If $I_1$ and $I_2$ are the magnitudes of induced current in the cases $I$ and $II$,respectively,then

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