The magnetic field in a region is given by $$\vec{B} = B_0\left(\frac{x}{a}\right)\hat{k}$$. A square loop of side $$d$$ is placed with its edges along the $$x$$ and $$y$$ axes. The loop is moved with a constant velocity $$\vec{v} = v_0\hat{i}$$. The emf induced in the loop is:

The magnetic field is $$\vec{B} = B_0\left(\frac{x}{a}\right)\hat{k}$$, which varies linearly with $$x$$. The square loop of side $$d$$ has its edges along the $$x$$ and $$y$$ axes, and it moves with velocity $$\vec{v} = v_0\hat{i}$$.

Let the left edge of the loop be at position $$x_0$$ at time $$t$$, so the right edge is at $$x_0 + d$$. The magnetic flux through the loop is $$\Phi = \int_0^d \int_{x_0}^{x_0+d} B_0\frac{x}{a}\,dx\,dy = d \cdot \frac{B_0}{a}\int_{x_0}^{x_0+d} x\,dx$$.

Evaluating the integral: $$\int_{x_0}^{x_0+d} x\,dx = \frac{(x_0+d)^2 - x_0^2}{2} = \frac{2x_0 d + d^2}{2} = x_0 d + \frac{d^2}{2}$$.

So $$\Phi = \frac{B_0 d}{a}\left(x_0 d + \frac{d^2}{2}\right)$$. Since $$x_0 = v_0 t + \text{const}$$, we have $$\frac{dx_0}{dt} = v_0$$.

The induced EMF is $$\varepsilon = -\frac{d\Phi}{dt} = -\frac{B_0 d}{a} \cdot d \cdot v_0 = -\frac{B_0 v_0 d^2}{a}$$.

The magnitude of the induced EMF is $$|\varepsilon| = \frac{B_0 v_0 d^2}{a}$$.