A rectangular loop of wire is placed in the X | vedclass

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$(C)$ The area of the loop is $A = 3 \,cm 	imes 4 \,cm = 12 \,cm^2 = 12 	imes 10^{-4} \,m^2$. The magnetic flux $\phi$ through the loop is $\phi = B

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$(C)$ The area of the loop is $A = 3 \,cm 	imes 4 \,cm = 12 \,cm^2 = 12 	imes 10^{-4} \,m^2$. The magnetic flux $\phi$ through the loop is $\phi = B \cdot A$. The induced emf $\varepsilon$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = -A \frac{dB}{dt}$. Since the loop is moving in a non-uniform magnetic field, the total rate of change of the magnetic field $B$ is given by the material derivative: $\frac{dB}{dt} = \frac{\partial B}{\partial t} + v \frac{\partial B}{\partial x}$. Given: $\frac{\partial B}{\partial t} = 2 	imes 10^{-2} \,T/s$ (increase in time). $\frac{\partial B}{\partial x} = -2 	imes 10^{-3} \,T/cm = -0.2 \,T/m$ (decrease along $X$-axis). $v = 10 \,cm/s = 0.1 \,m/s$. Substituting these values: $\frac{dB}{dt} = (2 	imes 10^{-2}) + (0.1) 	imes (-0.2) = 0.02 - 0.02 = 0 \,T/s$. Therefore, the induced emf $\varepsilon = -A 	imes 0 = 0 \,V$.

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