The back electromotive force in solenoid is $(u)$ a maximum when there is maximum rate 0 change of current. This occurs is in $A B$ part of the graph. So maximum back emf will be obtained between $5 \mathrm{~s}

Since, the back emf at $t=3 \mathrm{~s}$ is $e$

Also,

the rate of change of current at $t=3, \mathrm{~s}=$ slope of $O A$ from $t=0 \mathrm{~s}$ to $t=5 \mathrm{~s}=1 / 5 \mathrm{~A} / \mathrm{s}$.

So, we have

If $u=L 1 / 5\left(\right.$ for $\left.t=3 \mathrm{~s}, \frac{d I}{d t}=1 / 5\right)$ ( $L$ is a constant). Applying $\varepsilon=-L \frac{d I}{d t}$

Similarly, we have for other values

For $5 \mathrm{~s}

Thus, at $t=7 \mathrm{~s}, u_1=-3 \mathrm{e}$

For $10 \mathrm{~s}

$$u_2=L \frac{2}{20}=\frac{L}{10}=\frac{1}{2} e$$

For $t>30 \mathrm{~s}, u_2=0$

Thus, the back emf at $t=7 \mathrm{~s}, 15 \mathrm{~s}$ and 40 s are $-3 e, e / 2$ and 0 respectively.

Applying the mutual inductance of coil $A$ with respect to coil $B$

$$M_{21}=\frac{N_2 \phi_2}{I_1}$$

Therefore, we have

$$\text { Mutual inductance }=\frac{10^{-2}}{2}=5 \mathrm{mH}$$

Again applying this formula for other case

$$N_1 \phi_1=M_{12} I_2=5 \mathrm{mH} \times 1 \mathrm{~A}=5 \mathrm{mWb}$$

Let us assume that the parallel wires at are $y=0$ i.e., along $x$-axis and $y=d$. At $t=0, A B$ has $x=0$, i.e., along $y$-axis and moves with a velocity $v$. Let at time $t$, wire is at $x(t)=v t$.

Now, the motional emf across $A B$ is

$$=\left(B_0 \sin \omega t\right) v d(-\hat{\mathbf{j}})$$

emf due to change in field (along $O B A C$ )

$$=-B_0 \omega \cos \omega t x(t) d$$

Total emf in the circuit $=$ emf due to change in field (along $O B A C$ ) + the motional emf across $A B$

$$=-B_0 d[\omega x \cos (\omega t)+v \sin (\omega t)]$$

Electric current in clockwise direction is given by

$$=\frac{B_0 d}{R}(\omega x \cos \omega t+v \sin \omega t)$$

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

Substituting the values, we have

$$\begin{aligned} \text { Force needed along } \mathbf{i} & =\frac{B_0 d}{R}(\omega x \cos \omega t+v \sin \omega t) \times d \times B_0 \sin \omega t \\ & =\frac{B_0^2 d^2}{R}(\omega x \cos \omega t+v \sin \omega t) \sin \omega t \end{aligned}$$

This is the required expression for force./p>

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.