Question

A magnetic field $B = \frac{B_0}{x}$ exists in space along negative z direction. A square loop of side $L$ and resistance per unit length $R_0$ is moving with constant speed $v$ as shown. Initially side $PQ$ was nearly on y-axis. Find current in the loop at t = 2 sec.

Answer 1

$\frac{B_0 L}{8R_0 (2v+L)}$

Answer 2

The magnetic flux through the loop is $\Phi = \int_{x}^{x+L} \int_{0}^{L} -\frac{B_0}{x'} dy' dx' = -B_0 L \ln\left(\frac{x+L}{x}\right)$.

The induced EMF is $\mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d\Phi}{dx} \frac{dx}{dt}$. Given $\frac{d\Phi}{dx} = \frac{B_0 L^2}{x(x+L)}$ and $\frac{dx}{dt} = v$. So, $\mathcal{E} = -\frac{B_0 L^2 v}{x(x+L)}$. The magnitude is $|\mathcal{E}| = \frac{B_0 L^2 v}{x(x+L)}$.

The total resistance of the loop is $R = R_0 \times (4L)$. The current is $I = \frac{|\mathcal{E}|}{R} = \frac{B_0 L^2 v}{x(x+L)} \frac{1}{4R_0 L} = \frac{B_0 L v}{4R_0 x(x+L)}$.

At $t=2$ sec, $x = v \times 2 = 2v$. Substituting $x=2v$, $I = \frac{B_0 L v}{4R_0 (2v)(2v+L)} = \frac{B_0 L}{8R_0 (2v+L)}$.