(2)

The magnetic potential energy is given by

$V=-\overrightarrow{m}·\overrightarrow{B}=-mB\cos \theta$

when,    $\theta =90°;$  $V=-mB\cos 90°=0$

(1)

[IMAGE 228]----------------------------------

Let the length of the iron bar be $L$.
The magnetic moment of bar is given by

$M=mL$                                             ...(i)

Where, $m$ is pole strength.

When the bar is bent in the form of an arc,

$L=\frac{\theta }{360}\times 2\pi r=\frac{60°}{360°}\times 2\pi r$

$L=\frac{\pi r}{3}\Rightarrow r=\frac{3L}{\pi }$

From $∆PQS$ in the figure,

[IMAGE 229]----------------------------------------

$\sin 30°=\frac{x}{r}\Rightarrow x=\frac{r}{2}$

Distance between $PR=2x=r$

New magnetic moment

$M\text{'}=m\times r$

$\Rightarrow m\times \left(\frac{{\displaystyle 3L}}{{\displaystyle \pi }}\right); M\text{'}=\frac{3mL}{\pi }$

From equation (i), $M\text{'}=\frac{3M}{\pi }$

(3)

Magnetic moment, $M=1.0\times {10}^{-2}{\text{Am}}^2$

Moment of inertia, $I=\frac{{10}^{-6}}{{\pi }^2}$ $\text{kg}$ ${\mathrm{m}}^2$

Time period for one oscillation $\frac{10}{10}=1$$\text{s}$

We know that, $T=2\pi \sqrt{\frac{I}{MB}}$

$\therefore$    $T^2=\frac{4{\pi }^{2}I}{MB}$

$B=\frac{4{\pi }^{2}I}{M{T}^2}=\frac{4\times {(3.14)}^2}{1\times {10}^{-2}}\times \frac{{10}^{-6}}{{{\displaystyle (}{\displaystyle 3}{\displaystyle .}{\displaystyle 14}{\displaystyle )}}^2}=4\times {10}^{-4}=0.4$ $\text{mT}$

(4)

Time period of the needle, $T=\frac{5}{20}=\frac{1}{4}$$\text{s}$

Given, $I=9.8\times {10}^{-6}\text{kg}$ ${\mathrm{m}}^2, \mathrm{M}=\mathrm{x}\times {10}^{-5}$${\text{Am}}^2, \mathrm{B}=0.049$ $\text{T}$

Since, $T=2\pi \sqrt{\frac{I}{MB}}$

$\Rightarrow \frac{1}{4}=2\pi \sqrt{\frac{9.8\times {10}^{-6}}{x\times {10}^{-5}\times 0.049}}\Rightarrow \frac{1}{8\pi }=\sqrt{\frac{9.8\times {10}^{-6}}{x\times {10}^{-5}\times 0.049}}$

$\Rightarrow \frac{1}{64{\pi }^2}=\frac{9.8\times {10}^{-6}}{x\times {10}^{-5}\times 0.049}$

$\Rightarrow x=64{\pi }^2\times \frac{9.8\times {10}^{-1}}{0.049}=1280{\pi }^2$

(2)

[IMAGE 232]------------------------

$M_{\text{net}}=\sqrt{{\left(\frac{{\displaystyle M}}{{\displaystyle 2}}\right)}^2+{\left(\frac{M}{2}\right)}^2+2\frac{M}{2}·\frac{M}{2}\cos 120°}$

$M_{\text{net}}=\sqrt{\frac{M^2}{4}\times 2-\frac{M^2}{4}}=\sqrt{\frac{M^2}{4}}=\frac{M}{2}$

(4)

Work done in a coil

$W=mB(\cos {\theta }_1-\cos {\theta }_2)$

When it is rotated by angle $180°$, then

      $W=2mB=2\left(NIA\right)B$                                             ...(i)

Given,  $N=250, I=85\mu \text{A}=85\times {10}^{-6}\text{A}$

            $A=1.25\times 2.1\times {10}^{-4}{\text{m}}^2\approx 2.6\times {10}^{-4}{\text{m}}^2$

            $B=0.85$ $\text{T}$

Putting these values in eqn. (i), we get

        $W=2\times 250\times 85\times {10}^{-6}\times 2.6\times {10}^{-4}\times 0.85$

             $\approx 9.1\times {10}^{-6}\text{J}=9.1$ $\mu \text{J}$

(2)

At equilibrium, potential energy of dipole

   $U_i=-M{B}_H$

Final potential energy of dipole, $U_f=-M{B}_H\cos 60°=-\frac{M{B}_H}{2}$

$W=U_f-U_i=-\frac{M{B}_H}{2}-(-M{B}_H)=\frac{M{B}_H}{2}$                          ...(i)

             $\tau =M{B}_H\sin 60°=2W\times \frac{\sqrt{3}}{2} \text{[Using eqn. (i)]}$

             $=\sqrt{3}W$

(3)

The direction of magnetic dipole moment is from south to north pole of the magnet.

In configuration (1),

[IMAGE 234]---------------------------------

$m_{\text{net}}=\sqrt{m^2+m^2+2mm\cos 90°}=\sqrt{m^2+m^2}=m\sqrt{2}$

In configuration (2),

[IMAGE 235]---------------------------------

In configuration (3),

[IMAGE 236]---------------------------------

$m_{\text{net}}=\sqrt{m^2+m^2+2mm\cos 30°}=\sqrt{2m^2+2m^2\left(\frac{\sqrt{3}}{2}\right)}=m\sqrt{2+\sqrt{3}}$

In configuration (4),

[IMAGE 237]---------------------------------

$m_{\text{net}}=\sqrt{m^2+m^2+2mm\cos 60°}=\sqrt{2m^2+2m^2\left(\frac{1}{2}\right)}=m\sqrt{3}$