The magnetic field in a region is given by $B=B_0\left(1+\frac{x}{a}\right)\widehat{k}$ A square loop of edge-length 'd' is placed with its edges along the x and y-axes. The loop is moved with a constant velocity $\nu =v_0\widehat{i}$.  The emf induced in the loop is :

The magnetic field in a region is given by 

$B=B_0\left(1+\frac{x}{a}\right)\widehat{k}$ 

Emf can be given as,

$\epsilon =\overrightarrow{v}\times \overrightarrow{B}\cdot \overrightarrow{l}$

As, $\overrightarrow{v}\times \overrightarrow{B}$ is along the length of the conductor of the loop for two arms.

EMF produced in the arms,  

$\epsilon =\mathrm{vBI}$

Now for the remaining 2 arms, it will be perpendicular to length. Thus, emf in those arms = zero.

Suppose in some time, loop has moved a distance of x. So,

$\begin{array}{l}{\epsilon }_1=v_0\frac{B_0(x+d)}{a}d\\ {\epsilon }_2=v_0\frac{B_0{x}_d}{a}\end{array}$

Net emf,

$$\begin{gathered} {\epsilon }_{\text{net }}={\epsilon }_1-{\epsilon }_2 \\ =\frac{B_0{v}_0{d}^2}{a} \end{gathered}$$

Hence option (D) is correct.