1. Given magnetic field

   $\vec B = B_0\left(\frac{x}{a}\right)\hat k$

   So the magnetic field varies with the $x$-coordinate.

2. Loop orientation and motion

   • Square loop of side $d$
   • Its plane is in the $xy$-plane
   • Hence area vector is along $\hat k$
   • The loop moves with constant velocity $\vec v = v_0\hat i$

   Let the left edge of the square be at position $x$ at some instant. Then the right edge is at $x+d$.

3. Magnetic flux through the loop

   Magnetic flux is

   $\Phi = \iint \vec B\cdot d\vec A$

   Since $\vec B$ is along $\hat k$ and $d\vec A = \hat k\,dx\,dy$,

   $\Phi = \int_0^d \int_x^{x+d} B_0\left(\frac{x'}{a}\right) dx'\,dy$

   where $x'$ is the integration variable.

   First integrate over $y$:

   $\Phi = d\int_x^{x+d} B_0\left(\frac{x'}{a}\right)dx'$

   $\Phi = \frac{B_0 d}{a}\int_x^{x+d} x'\,dx'$

   $\Phi = \frac{B_0 d}{a}\left[\frac{x'^2}{2}\right]_x^{x+d}$

   $\Phi = \frac{B_0 d}{2a}\left((x+d)^2-x^2\right)$

   $\Phi = \frac{B_0 d}{2a}(2xd+d^2)$

   $\Phi = \frac{B_0 d^2}{a}x + \frac{B_0 d^3}{2a}$

4. Induced emf

   Using Faraday's law,

   $\mathcal E = \left|\frac{d\Phi}{dt}\right|$

   Since the loop moves with speed $v_0$ along $+x$,

   $\frac{dx}{dt} = v_0$

   Therefore,

   $\frac{d\Phi}{dt} = \frac{B_0 d^2}{a}\frac{dx}{dt} = \frac{B_0 v_0 d^2}{a}$

   Hence,

   $\boxed{\mathcal E = \frac{B_0 v_0 d^2}{a}}$

5. Check options

   • A: $\dfrac{B_0 v_0 d^2}{2a}$ → incorrect
   • B: $\dfrac{B_0 v_0^2 d}{2a}$ → incorrect dimensions
   • C: $\dfrac{B_0 v_0 d}{2a}$ → incorrect
   • D: $\dfrac{B_0 v_0 d^2}{a}$ → correct

So the correct option is D.