Magnetism

From right hand rule, we write, the force on a current carrying wire placed in a magnetic field is given by $\overrightarrow{\mathrm{F}}=(l \times \mathrm{B}) \hat{i}$

If $\phi$ is angle between $l$ and B , then $\overrightarrow{\mathrm{F}}=i l \mathrm{~B} \sin \phi$ Force on $\mathrm{AB}=\mathrm{BIL}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_1} \mathrm{IL}_{\mathrm{AB}}$.

$ \mathrm{F}_{\mathrm{AB}}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_1} \mathrm{I}\left(\frac{\phi}{2 \pi} 2 \pi r_1\right)=\frac{\mu_0 \mathrm{I} \mathrm{I}_1 \phi}{2 \pi r} $

(This is inward force)

Force on $\mathrm{CD}=\mathrm{BIL}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_2} \mathrm{IL}_{\mathrm{CD}}$.

$ \mathrm{F}_{\mathrm{CD}}=\frac{\mu_0 \mathrm{I}_1}{2 \pi r_2}\left(\frac{\phi}{2 \pi} 2 \pi r_2\right)=\frac{\mu_0 \mathrm{II}_1 \phi}{2 \pi} $

(This is out wards)

Here, net force is zero.

$B C$ and $C D$ are radius vectors. The $I$ and $B$ are $\perp$ er here,

$ \therefore=i l \mathrm{~B} \sin 90^{\circ}=i l \mathrm{~B} . $

The force acting on the both section is same as the $y$ experience same magnetic field at every point on the wire.

But the direction is different for both of them.

$\therefore$ Force on BC acts upwards and AD acts downwards.

Thus it appears rotating clockwise as scene from O .

(A) Any static charge produces electric field $\forall$ invariant with time. Uniformly charged dielectric does the same.

(B) Rotating charged behaves like a current. hence, magnetic field is produced around the ring and also if there is current in rotating ring, ring has magnetic moment.

(C) Since, net charge is zero there will be no time independent electric field. The current produces magnetic field and magnetic moment.

(D) A changing magnetic field will be produced. This will create a induced electric field.

Also a changing magnetic moment will be produced.

(A) The question is based on the theory of a moving coil galvanometer.

Torque on Coil

$\begin{aligned} \tau & =\mathrm{MB} \sin \theta \\ \tau & =(\mathrm{N} i \mathrm{~A}) \mathrm{B} \sin 90^{\circ} \\ \tau & =\mathrm{N} i \mathrm{AB} \\ k i & =\mathrm{NiAB} \\ k & =\mathrm{NAB} \end{aligned}$

(B) The torsional constant of the spring is defined as the torque per unit twist. Thus $\tau= \mathrm{C} \theta$.

But

$iNAB = C\theta $

Given, $i = {i_0}$ and $\theta = \pi /2$

${i_0}NAB = C{\pi \over 2}$

Hence, the torsional constant of the spring

$C = {{2N{i_0}AB} \over \pi }$

(C) Angular impulse $=\int \tau d t=\int \mathrm{NiAB} d t$

$=\mathrm{NAB} \int i d t=\mathrm{NABQ}$

But, I $\omega=\mathrm{NABQ}$ (or) $\omega=\frac{\text { NABQ }}{\mathrm{I}}$

$\therefore$ Rotational kinetic energy

$\begin{aligned} & =\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{NABQ}}{\mathrm{I}}\right)^{2} \\ & =\frac{1}{2} \times \frac{\mathrm{N}^{2} A^{2} \mathrm{~B}^{2} Q^{2}}{\mathrm{I}} \end{aligned}$

$\therefore$ Potential energy $=\frac{1}{2} \mathrm{CQ}_{\max }^{2}$

On equating both Kinetic and Potential energy

$\begin{aligned} \frac{1}{2} \mathrm{Q}_{\max }^{2} \mathrm{C} & =\frac{1}{2} \times \frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} \mathrm{Q}^{2}}{\mathrm{I}} \\ \mathrm{Q}_{\max }^{2} & =\frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} a^{2}}{\mathrm{I}} \times \frac{1}{\mathrm{C}} \\ & =\frac{\mathrm{N}^{2} \mathrm{~A}^{2} \mathrm{~B}^{2} \mathrm{Q}^{2}}{\mathrm{I}} \times \frac{\pi}{2 i_{0} \mathrm{NAB}} \\ & =\frac{\mathrm{Q}^{2} \mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}} \\ \mathrm{Q}_{\max } & =\sqrt{\frac{\mathrm{Q}^{2} \mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}}}=\mathrm{Q} \sqrt{\frac{\mathrm{NAB} \pi}{2 \mathrm{Ii}_{0}}} \end{aligned}$