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(A) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.The magnetic flux $\phi$ thr

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(A) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.The magnetic flux $\phi$ through the square loop is $\phi = \int B \cdot dA = \int_{x}^{x+a} \frac{\mu_0 I}{2\pi r} (a \, dr) = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}ight)$.The mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}ight)$.Since the loop is moving at a constant speed $v$,the distance $x$ increases with time $t$ as $x = x_0 + vt$.Substituting this into the expression for $M$,we get $M(t) = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x_0 + vt}ight)$.As $t$ increases,the term $\frac{a}{x_0 + vt}$ decreases,and therefore $\ln(1 + \frac{a}{x_0 + vt})$ decreases.This corresponds to a curve that starts at a maximum value and decreases asymptotically towards zero as $t 	o \infty$,which matches the graph shown in option $A$.

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