The magnetic field is $\mathbf{B} = B_y \hat{j}$. When it is switched off, $\frac{\partial \mathbf{B}}{\partial t}$ is in the $-\hat{j}$ direction. This change induces an electric field $\mathbf{E}$.

From Maxwell’s equation $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$:

$$ \frac{\partial E_x}{\partial z} – \frac{\partial E_z}{\partial x} = -\frac{\partial B_y}{\partial t} $$

Assuming suitable symmetry where $E_z = 0$, we get $\frac{\partial E_x}{\partial z} = -\dot{B}$. Integrating this with respect to $z$ (setting $E=0$ at $z=0$):

$$ E_x = -z \frac{dB}{dt} $$

The box has charge $+Q$ on the top face ($z = +h/2$) and $-Q$ on the bottom face ($z = -h/2$), where $Q = \sigma a b$.

Force on Top Plate:

$$ F_{top} = Q E_x(h/2) = (\sigma ab) \left( -\frac{h}{2} \dot{B} \right) $$

Force on Bottom Plate:

$$ F_{bottom} = (-Q) E_x(-h/2) = (-\sigma ab) \left( – \left(-\frac{h}{2}\right) \dot{B} \right) = (\sigma ab) \left( -\frac{h}{2} \dot{B} \right) $$

Total Force:

$$ F_{net} = F_{top} + F_{bottom} = -\sigma ab h \frac{dB}{dt} \hat{i} $$

Impulse $\mathbf{J} = \int \mathbf{F}_{net} dt$.

$$ \mathbf{J} = \int -\sigma ab h \frac{dB}{dt} \hat{i} \, dt = -\sigma ab h \hat{i} \int_{B_0}^{0} dB $$

$$ \mathbf{J} = -\sigma ab h \hat{i} (0 – B_0) = \sigma ab h B_0 \hat{i} $$

The velocity acquired is $\mathbf{v} = \frac{\mathbf{J}}{m}$:

$$ \mathbf{v} = \frac{\sigma a b h B_0}{m} \hat{i} $$