Consider the parallel wires along $\mathrm{x}$-axis at $\mathrm{y}=0$ and $\mathrm{y}=l$, at $t=0$, so $X Y$ has $x=0$ i.e., along $y$-axis.
(i) Let the wire $X Y$ beat $x=x(t)$ at time $t=t$, and at $t=0$ is at $\mathrm{x}=0$,
The magnetic flux linked with the loop is
$$
\phi=\text { B.A }=\text { BA } \cos 0=\text { BA }
$$
Magnetic flux at any instant $t$ time is
$$
\phi(t)=B(t)[l \times x(t)]
$$
Total emf in the circuit $=$ emf due to change in field (along XYAC) + the motional emf across XY
$$
\begin{aligned}
&\mathrm{E}=\frac{-\mathrm{d} \phi(\mathrm{t})}{\mathrm{dt}}=\frac{-\mathrm{dB}(\mathrm{t})}{\mathrm{dt}} l \mathrm{x}(\mathrm{t})-\mathrm{B}(\mathrm{t}) l\left(\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}\right) \\
&\mathrm{E}=-\frac{\mathrm{d} \phi(\mathrm{t})}{\mathrm{dt}}=-\frac{\mathrm{dB}(\mathrm{t})}{\mathrm{dt}} l \mathrm{x}(\mathrm{t})-\mathrm{B}(\mathrm{t}) l v(\mathrm{t})
\end{aligned}
$$
By Fleming right hand rule :
Electric current in clockwise direction is
$$
I=\frac{E}{R} \alpha
$$
The force acting on the conductor is
$$
\mathrm{F}=\mathrm{i} / \mathrm{B} \sin 90^{\circ}=\mathrm{i} / \mathrm{B}(\mathrm{t})
$$
By putting all the values in above equation, we get Force $=\frac{I B(t)}{R}\left[-\frac{d B(t)}{d t} \operatorname{Ix}(t)-B(t) \operatorname{Iv}(t)\right] \hat{i}$ From Newton's second law of motion, which is the required equation.


(ii) If $\mathrm{B}$ is independent of time i.e., $\mathrm{B}=$ Constant ( $B$ does not change with time)
$$
\therefore \quad \begin{aligned}
&\frac{\mathrm{dB}}{\mathrm{dt}}=0 \\
&\mathrm{~B}(\mathrm{t})=\mathrm{B} \text { and } v(\mathrm{t})=\mathrm{v}
\end{aligned}
$$
Substituting the above value in Eq (i), we get,
$$
\begin{aligned}
&\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt^{2 }}}=\frac{-\mathrm{I}^2}{\mathrm{mR}}[\mathrm{B}+\mathrm{Bv}] \\
&\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{I}^2 \mathrm{~B}^2}{\mathrm{mR}} \mathrm{v}=0
\end{aligned}
$$
Integrating using variable separable form of differential equation, we have
$$
v=\operatorname{Aexp}\left(\frac{-I^2 B^2 t}{m R}\right)
$$
Now, from given conditions, at $\mathrm{t}=0, \mathrm{v}=\mathrm{u}_0$ $v(t)=u_0 \exp \left(-\mathrm{I}^2 \mathrm{~B}^2 t / \mathrm{mR}\right)$
Which is the required equation,
(iii) As we know that, the power consumption is
$$
P=I^2 R
$$
Here, $\quad I^2 R=\frac{B^2 I^2 v^2(t)}{R^2} \times R \quad\left[\therefore I=\frac{B I V t}{R}\right]$
Power loss $=\frac{\mathrm{B}^2 \mathrm{I}^2}{\mathrm{R}} \mathrm{u}_0^2 \exp \left(-2 \mathrm{I}^2 \mathrm{~B}^2 \mathrm{t} / \mathrm{mR}\right)$ So, energy consumed in time interval $\mathrm{dt}$ is $=\mathrm{Pdt}=\mathrm{I}^2 \mathrm{Rdt}$
Then, the total energy consumed in time $t$
$$
\begin{gathered}
\quad=\int_0^{\mathrm{t}} \mathrm{I}^2 \mathrm{Rdt} \\
=\frac{\mathrm{B}^2 \mathrm{I}^2}{\mathrm{R}} \mathrm{u}_0^2 \frac{\mathrm{mR}}{2 \mathrm{I}^2 \mathrm{~B}^2}\left[1-\mathrm{e}^{-2\left(\mathrm{I}^2 \mathrm{~B}^2 / \mathrm{mr}\right)}\right] \\
=\frac{\mathrm{mu}_0^2}{2}\left[1-\mathrm{e}^{\left(-21^2 \mathrm{~B}^2 t / \mathrm{mr}\right)}\right]
\end{gathered}
$$
So, total energy consumed in time (t) is $=$ (initial K.E.) $-$ (Final K.E.)
$$
=\left[\frac{\mathrm{m}}{2} \mathrm{u}_0^2-\frac{\mathrm{m}}{2} \mathrm{v}^2(\mathrm{t})\right]
$$