For equilibrium/ balance, net torque should also be equal to zero.

When the field is off $\sum t=0$ considering the separation of each hung from mid-point be $I$.

$$\begin{aligned} M g l & =W_{\text {coil }} l \\ 500 \mathrm{gl} & =W_{\text {coil }} l \\ W_{\text {coil }} & =500 \times 9.8 \mathrm{~N} \end{aligned}$$

Taking moment of force about mid-point, we have the weight of coil

When the magnetic field is switched on

$$\begin{aligned} M g l+m g l & =W_{\text {coil }} l+I B L \sin 90 Y I \\ m g l & =\text { BIL } l \\ m & =\frac{B I L}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g} \end{aligned}$$

Thus, 1 g of additional mass must be added to regain the balance.

The thicker wire has a resistance $R$, then the other wire has a resistance $2 R$ as the wires are of the same material but with cross-sections differing by a factor 2.

Now, the force and hence, torque on first wire is given by

$$F_1=i_1 l B=\frac{V_0}{2 R} l B, \tau_1=\frac{d}{2 \sqrt{2}} F_1=\frac{V_0 l d B}{2 \sqrt{2} R}$$

Similarly, the force hence torque on other wire is given by

$$F_2=i_2 l B=\frac{V_0}{2 R} l B, \tau_2=\frac{d}{2 \sqrt{2}} F_2=\frac{V_0 l d B}{4 \sqrt{2} R}$$

$$\begin{aligned} &\text { So, net torque, }\\ &\begin{aligned} \tau & =\tau_1-\tau_2 \\ \tau & =\frac{1}{4 \sqrt{2}} \frac{V_0 l d B}{R} \end{aligned} \end{aligned}$$

Since, $B$ is along the $x$-axis, for a circular orbit the momenta of the two particles are in the $y$-z plane. Let $p_1$ and $p_2$ be the momentum of the electron and positron, respectively. Both traverse a circle of radius $R$ of opposite sense. Let $p_1$ make an angle $\theta$ with the $y$-axis $p_2$ must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance $R$. Let the centre of the electron be at $C_e$ and of the positron at $C_p$. The coordinates of $C_e$ is

$$C_e \equiv(0,-R \sin \theta, R \cos \theta)$$

The coordinates of $C_p$ is

$$C_p \equiv\left(0,-R \sin \theta, \frac{3}{2} R-R \cos \theta\right)$$

The circles of the two shall not overlap if the distance between the two centers are greater than $2 R$.

Let $d$ be the distance between $C_p$ and $C_e$.

Let $d$ be the distance between $C_p$ and $C_e$.

Then,

$$\begin{aligned} d^2 & =(2 R \sin \theta)^2+\left(\frac{3}{2} R-2 R \cos \theta\right)^2 \\ & =4 R^2 \sin ^2 \theta+\frac{9^2}{4} R-6 R^2 \cos \theta+4 R^2 \cos ^2 \theta \\ & =4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta \end{aligned}$$

Since, $d$ has to be greater than $2 R$

$$\begin{array}{rlrl} & d^2 >4 R^2 \\ \Rightarrow & 4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta >4 R^2 \\ \Rightarrow & \frac{9}{4} >6 \cos \theta \\ \text { or, } & \cos \theta <\frac{3}{8} \end{array}$$

We know that magnetic moment of the coils $m=n I A$.

Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.

(i) for an equilateral triangle of side $a$, $n=4$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n I A=4 I\left(\frac{\sqrt{3}}{4} a^2\right)$

$$\therefore \quad m=I a^2 \sqrt{3}$$

(ii) For a square of sides $a$,

$A=a^2$

$n=3$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n I A=3 I\left(a^2\right)=3 I \mathrm{a}^2$

(iii) For a regular hexagon of sides a,

$n=2$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n / A=21\left(\frac{6 \sqrt{3}}{4} a^2\right)$

$$m=3 \sqrt{3} a^2 I$$

$m$ is in a geometric series.

(a) B (z) points in the same direction on $z$-axis and hence, $J(L)$ is a monotonically function of L.

Since, $B$ and $\mathbf{d l}$ along the same direction, therefore $\mathbf{B} . \mathbf{d l}=\mathbf{B} \cdot \mathrm{dl}$ as $\cos 0=1$

(b) $J(L)+$ contribution from large distance on contour $C=\propto_0 I$

$$\begin{aligned} & \therefore \quad \text { as } L \rightarrow \infty \\ & \text { Contribution from large distance } \rightarrow 0\left(\text { as B } \propto 1 / r^3\right) \\ & \qquad J(\infty)-\alpha_0 I \end{aligned}$$

(c) The magnetic field due to circular current-carrying loop of radius $R$ in the $x$-y plane with centre at origin at any point lying at a distance of from origin.

$$ \begin{aligned} B_z & =\frac{\alpha_0 I R^2}{2\left(z^2+R^2\right)^{3 / 2}} \\ \int_{-\infty}^{\infty} B_z d z & =\int_{-\infty}^{\infty} \frac{\alpha_0 I R^2}{2\left(z^2+R^2\right)^{3 / 2}} d z \\ \text{Put}\quad z & =R \tan \theta_1 \\ \Rightarrow\quad d z & =R \sec ^2 \theta d \theta \\ \therefore\quad\int_{-\infty}^{\infty} B_z d z & =\frac{\propto_0 I}{2} \int_{-\pi / 2}^{\pi / 2} \cos \theta d \theta=\propto_0 I \end{aligned}$$

(d) $B(z)_{\text {square }}

$$\therefore \quad \Im(L)_{\text {square }}<\Im(L)_{\text {circular coil }}$$

But by using arguments as in (b)

$$\Im(\infty)_{\text {square }}=\Im(\infty)_{\text {circular }}$$