Physics M.C.Q (1 Marks)

$B$. If sheet is non-magnetic, no force needed.

$C$. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.

$D$. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

$M=m I$.

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$\Delta l=2 \frac{l}{2} \sin 30^{\circ}$

$=\frac{l}{2}$

$M^{\prime}=m / / 2$

$=M / 2$

Magnetic monopole doesn't exist.

Hence net magnetic flux through any closed surface is zero.

$\mu=\mu_{ o } \mu_{ r }=4 \pi \times 10^{-7} \times 600$

$=2.4 \pi \times 10^{-4} \frac{ Tm }{ A }$

$\mathrm{H}=\mathrm{B}_{\mathrm{E}} \cos \delta$

$W=m B\left(\cos \theta_{1}-\cos \theta_{2}\right)$

When it is rotated by angle $180^{o}$ then

$W=2 m B=2(N I A) B$                   .......... $(i)$

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

$A=1.25 \times 2.1 \times 10^{-4}\, \mathrm{m}^{2}=2.5 \times 10^{-4}\, \mathrm{m}^{2}$

$B=0.85\, \mathrm{T}$

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

$W= 2 \times 250 \times 85 \times 10^{-6} \times 2.5 \times 10^{-4} \times 0.85$

$=9.1 \times 10^{-6} \,\mathrm{J}=9.1\, \mu \mathrm{J}$

$\tan \theta = \frac{B_V}{B_H}$

 or $cot\theta = \frac{B_H}{B_V}$

Suppose planes $1$ and $2$ are two mutually perpendicular planes and respectively make angles $\theta$ and $90°- \theta$ with the magnetic meridian. The vertical components of earth's magnetic field remain same in the two planes but the effective horizontal components in the planes will be

$B_{1}=B_{H} \cos \theta \text { and } B_{2}=B_{H} \sin \theta$

The angles of $\operatorname{dip} \theta_{1}$ and $\theta_{2}$ in the two planes are given by

$\tan \theta_{1}=\frac{B_{V}}{B_{1}}$

$\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \theta}$

or $\quad \cot \theta_{1}=\frac{B_{H} \sin \theta}{B_{V}}.........(ii)$

$\text { Similarly, } \cot \theta_{2}=\frac{B_{H} \sin \theta}{B_{V}}.........(iii)$

From eqns. $(ii)$ and $(iii)$

$\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{B_{H}^{2}}{B_{V}^{2}}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{B_{H}^{2}}{B_{V}^{2}}$

$\therefore \quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta \quad[\text { from eqn. (i) }]$

$U_{i}=-M B_{H}$

Final potential energy of dipole,

${U_{f}=-M B_{H} \cos 60^{\circ}=-\frac{M B_{H}}{2}}$

$W = {U_f} - {U_i} =  - \frac{{M{B_H}}}{2} - \left( { - M{B_H}} \right) = \frac{{M{B_H}}}{2} \ldots ...(i)$

Required torque, $\tau=M B_{H} \sin 60^{\circ}$

$\tau=2 W \times \frac{\sqrt{3}}{2} \quad[\text { Using eqn. ( i ) }]$

$=\sqrt{3}\,W$

$\therefore $New magnetic moment $M' = \frac{m}{2} \times l = \frac{{ml}}{2} = \frac{M}{2}$

If cut perpendicular to the axis of magnet, then new pole strength $m' = m$ and new length, $l' = l/2$

$\therefore $ New magnetic moment $M' = m \times \frac{l}{2} = \frac{{ml}}{2} = \frac{M}{2}$

Magnetic moment of each part

$ = M' = m'L' = mL = \frac{M}{2}$

$|\overrightarrow{\mathrm{F}}|=\mathrm{mg}$

$\therefore \frac{3 \mu_{0}}{2 \pi r^{4}}\left(\mathrm{m}_{\mathrm{b}} \mathrm{m}_{\mathrm{a}}\right)=\mathrm{mg}$

$r=\left(\frac{3 \mu_{0} m^{2}}{2 \pi m g}\right)^{1 / 4}$

where ${{\text{B}}_1} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2{\text{M}}}}{{{{\text{d}}^3}}}\,and\,{{\text{B}}_2} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{\text{M}}}{{{{\text{d}}^3}}}$

$ \Rightarrow {{\text{B}}_{{\text{rest}}}} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{\sqrt 5 {\text{M}}}}{{{{\text{d}}^3}}}$

$\mathrm{B}_{\text {equatorial }}=\frac{\mathrm{M}}{\left(\frac{\mathrm{x}}{2}\right)^{32}}=\frac{\mathrm{M}}{\mathrm{x}^{3}}$.........$(ii)$

From equation $(i)$ and $(ii)$, $B$$_{\text {equatorial }}=36$ $gauss.$

$\mathrm{B}_{\text {equi }}=-\frac{\mu_{0}}{4 \mathrm{n}} \frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{r}^{3}}$

if $\theta \neq 0$ or $180^{\circ}$

so $\mathrm{PQ}_{1}$ and $\mathrm{PQ}_{2}$

$4 \times 10^{-5}=\frac{4 \pi \times 10^{-7} \times M}{4 \pi \times\left(6.4 \times 10^{3}\right)^{3}}$

$M=1.05 \times 10^{23}\, \mathrm{A}-\mathrm{m}^{2}$

Accordy to fleming left hand rule.

$T=2 \pi \sqrt{\frac{I}{M B}}$

$T^{\prime}=2 \pi \sqrt{\frac{I}{n^3 \frac{M}{n} B}}=\frac{2 \pi}{n} \sqrt{\frac{I}{M B}}$

$T^{\prime}=\frac{T}{n}$