Let us assume that the parallel wires at are $y=0$, i.e., along $x$-axis and $y=l$. At $t=0, X Y$ has $x=0$ i.e., along $y$-axis.

(i) Let the wire be at $x=x(t)$ at timet .

The magnetic flux linked with the loop is given by

$$\begin{aligned} & \qquad \phi=\mathbf{B} \cdot \mathbf{A}=B A \cos 0=B A \\ & \text { at any instant } t \quad \text { Magnetic fluX }=B(t)(l \times x(t)) \\ & \text { Total emf in the circuit }=\text { emf due to change in field (along } X Y A C)+ \text { the motional emf } \\ & \text { across } X Y \end{aligned}$$

$E=-\frac{d \phi}{d t}=-\frac{d B(t)}{d t} l x(t)-B(t) l v(t) \quad$ [second term due to motional emf]

Electric current in clockwise direction is given by

$$I=\frac{1}{R} E$$

The force acting on the conductor is given by $F=i l B \sin 90 \Upsilon=i l B$

Substituting the values, we have

$$\text { Force }=\frac{I B(t)}{R}\left[-\frac{d B(t)}{d t} I x(t)-B(t) I v(t)\right] \hat{\mathbf{i}}$$

Applying Newton's second law of motion,

$$m \frac{d^2 x}{d t^2}=-\frac{I^2 B(t)}{R} \frac{d B}{d t} x(t)-\frac{I^2 B^2(t)}{R} \frac{d x}{d t}\quad\text{.... (i)}$$

which is the required equation.

(ii) If $\mathbf{B}$ is independent of time i.e., $B=$ Constant Or

$$\frac{d B}{d t}=0$$

Substituting the above value in Eq (i), we have

$$\begin{aligned} \frac{d^2 x}{d t^2}+\frac{I^2 B^2}{m R} \frac{d x}{d t} & =0 \\ \text{or}\quad\frac{d v}{d t}+\frac{I^2 B^2}{m R} v & =0 \end{aligned}$$

Integrating using variable separable form of differential equation, we have

$$v=A \exp \left(\frac{-I^2 B^2 t}{m R}\right)$$

Applying given conditions,

$$\begin{aligned} & \text { at } t=0, v=u_0 \\ & v(t)=u_0 \exp \left(-I^2 B^2 t / m R\right) \end{aligned}$$

This is the required equation.

(iii) Since the power consumption is given by $P=I^2 R$

Here,

$$\begin{aligned} I^2 R & =\frac{B^2 I^2 V^2(t)}{R^2} \times R \\ & =\frac{B^2 I^2}{R} u_0^2 \exp \left(-2 I^2 B^2 t I m R\right) \end{aligned}$$

Now, energy consumed in time interval $d t$ is given by energy consumed $=P d t=I^2 R d t$

Therefore, total energy consumed in time $t$

$$\begin{aligned} & \left.=\int_0^t I^2 R d t=\frac{B^2 I^2}{R} u_0^2 \frac{m R}{2 I^2 B^2}\left[1-e^{-\left(l^2 B^2 t / m r\right.}\right)\right] \\ & =\frac{m}{2} u_0^2-\frac{m}{2} v^2(t) \\ & =\text { decrease in kinetic energy. } \end{aligned}$$

This proves that the decrease in kinetic energy of $X Y$ equals the heat lost in $R$.

Let us consider the position of rotating conductor at time interval

$$t=0 \text { to } t=\frac{\pi}{4 \omega}(\text { or } T / 8)$$

the rod $O P$ will make contact with the side $B D$. Let the length $O Q$ of the contact at sometime $t$ such that $0

$$\begin{aligned} \phi & =B \frac{1}{2} Q D \times O D=B \frac{1}{2} l \tan \theta \times l \\ & =\frac{1}{2} B i^2 \tan \theta, \text { where } \theta=\omega t \end{aligned}$$

Applying Faraday's law of EMI,

Thus, the magnitude of the emf generated is $\varepsilon=\frac{d \phi}{d t}=\frac{1}{2} B l^2 \omega \sec ^2 \omega t$

The current is $I=\frac{\varepsilon}{R}$ where $R$ is the resistance of the rod in contact. where, $R \propto \lambda$

$$\begin{aligned} & R=\lambda x=\frac{\lambda l}{\cos \omega t} \\ \therefore\quad & I=\frac{1}{2} \frac{B l^2 \omega}{\lambda l} \sec ^2 \omega t \cos \omega t=\frac{B l \omega}{2 \lambda \cos \omega t} \end{aligned}$$

Let the length $O Q$ of the contact at some timet such that $\frac{\pi}{4 \omega}

$$\begin{aligned} & \phi=\left(l^2+\frac{1}{2} \frac{l^2}{\tan \theta}\right) B \\ \text{where,}\quad & \theta=\omega t \end{aligned}$$

Thus, the magnitude of emf generated in the loop is

$\varepsilon=\frac{d \phi}{d t}=\frac{1}{2} B l^2 \omega \frac{\sec ^2 \omega t}{\tan ^2 \omega t}$

The current is $I=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{\varepsilon \sin \omega t}{\lambda l}=\frac{1}{2} \frac{B l \omega}{\lambda \sin \omega t}$

Similarly for $\frac{3 \pi}{4 \omega} < t < \frac{\pi}{\omega}$ or $\frac{3 T}{8} < t < \frac{T}{2}$, the rod will be in touch with $A C$.

The flux through $O Q A B D$ is given by

$$\phi=\left(2 l^2-\frac{l^2}{2 \tan \omega t}\right) B$$

And the magnitude of emf generated in loop is given by

$$\begin{aligned} & \varepsilon=\frac{d \phi}{d t}=\frac{B \omega l^2 \sec ^2 \omega t}{2 \tan ^2 \omega t} \\ & l=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{1}{2} \frac{B l \omega}{\lambda \sin \omega t} \end{aligned}$$

These are the required expressions.

Let us consider a strip of length $l$ and width $d r$ at a distance $r$ from infinite long current carrying wire. The magnetic field at strip due to current carrying wire is given by

Field $B(r)=\frac{\propto_0 I}{2 \pi r}$ (out of paper)

Total flux through the loop is

$$\text { Flux }=\frac{\propto_0 I}{2 \pi} l \int_{x_0}^x \frac{d r}{r}=\frac{\propto_0 I}{2 \pi} \ln \frac{x}{x_0}\quad\text{.... (i)}$$

The emf induced can be obtained by differentiating the eq. (i) wrt $t$ and then applying Ohm's law

$$\begin{aligned} &\frac{\varepsilon}{R}=I\\ &\text { We have, induced current }=\frac{1}{R} \frac{d \phi}{d t}=\frac{\varepsilon}{R}=\frac{\propto_0 I}{2 \pi} \frac{\lambda}{R} \ln \frac{x}{x_0}\quad\left(\because \frac{d \mathrm{I}}{d t}=\lambda\right) \end{aligned}$$

The emf induced can be obtained by differentiating the expression of magnetic flux linked wrtt and then applying Ohm's law

$$I=\frac{E}{R}=\frac{1}{R} \frac{d \phi}{d t}$$

We know that electric current

$$I(t)=\frac{d Q}{d t} \text { or } \frac{d Q}{d t}=\frac{1}{R} \frac{d \phi}{d t}$$

Integrating the variable separable form of differential equation for finding the charge $Q$ that passed in time $t$, we have

$$\begin{aligned} Q\left(t_1\right)-Q\left(t_2\right) & =\frac{1}{R}\left[\phi\left(t_1\right)-\phi\left(t_2\right)\right] \\ \phi\left(t_1\right) & =L_1 \frac{\alpha_0}{2 \pi} \int_x^{L_2+x} \frac{d x^{\prime}}{x^{\prime}} I\left(t_1\right) \quad\text{[Refer to the Eq. (i) of answer no.25] }\\ & =\frac{\alpha_0 L_1}{2 \pi} I\left(t_1\right) \ln \frac{L_2+x}{x} \end{aligned}$$

The magnitude of charge is given by,

$$\begin{aligned} & =\frac{\propto_0 L_1}{2 \pi} \ln \frac{L_2+x}{x}\left[I_0+0\right] \\ & =\frac{\propto_0 L_1}{2 \pi} I_1 \ln \left(\frac{L_2+x}{x}\right) \end{aligned}$$

This is the required expression.

Since, the magnetic field is brought to zero in time $\Delta t$, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring by the phenomenon of EMI. The induces emf causes the electric field $E$ generation around the ring.

The induced emf $=$ electric field $E \times(2 \pi b)($ Because $V=E \times d)$ ..... (i)

By Faraday's law of EMI

The induced emf $=$ rate of change of magnetic flux

$=$ rate of change of magnetic field $\times$ area

$$=\frac{B л a^2}{\Delta t}\quad\text{... (ii)}$$

From Eqs. (i) and (ii), we have

$$2 \pi b E=\mathrm{e} m f=\frac{B \pi a^2}{\Delta t}$$

Since, the charged ring experienced a electric force $=Q E$ This force try to rotate the coil, and the torque is given by

$$\begin{aligned} \text { Torque } & =b \times \text { Force } \\ & =Q E b=Q\left[\frac{B \pi a^2}{2 \pi b \Delta t}\right] b \\ & =Q \frac{B a^2}{2 \Delta t} \end{aligned}$$

If $\Delta L$ is the change in angular momentum

$$\Delta L=\text { Torque } \times \Delta t=Q \frac{B a^2}{2}$$

Since, initial angular momentum $=0$

Now, since $\quad$ Torque $\times \Delta t=$ Change in angular momentum

Final angular momentum $=m b^2 \omega=\frac{Q B a^2}{2}$

$$\omega=\frac{Q B a^2}{2 m b^2}$$

On rearranging the terms, we have the required expression of angular speed.