Magnetism And Matter – JEE Main PhysicsPractice Questions & PYQs

The relation between magnetisation and magnetising field, $\text{M}=\chi \text{H}$ For diamagnetic material, the magnetic susceptibility $(\chi )$ is negative. The correct graph is (A) and (C)

If apparent dip circle is at an angle α with true dip circle then As $\cos \alpha < 1$ Hence true dip $(\phi )$ is less than apparent ${\text{dip}}(\phi ')$

${T}=2 \pi \\sqrt{\frac{{I}}{{MB}}}$ ${B}=80 \\times 10^{-4}=8 {mT}$

Statement (A) Magnetic monopoles do not exist, they always exist in pairs as South and North poles but electric monopoles exist. So, statement ${A}$ is incorrect. Statement (B) In solenoid have a both ends, so magnetic field lines cannot be confined. Inside the solenoids, the magnetic field lines are parallel to the length of the conductor. So, statement B is incorrect. Statement (C) In toroid has no ends. It is closed loop conductor, so the magnetic field lines are completely confined. So, the statement $C$ is correct. Statement (D) Inside the bar magnet, the magnetic field lines are parallel. So, the statement D is incorrect. Statement (E) Magnetic susceptibility of the diamagnetic material is $-1$. So, the statement $E$ is correct. Hence, the correct statements are $C$ and $E$.

As temperature increases, domains disintegrate so ferromagnetism decreases and above curie temperature it become paramagnet.

The given situation can be shown as belowFrom the above figure, it is clear that the neutral point will lie on equitorial plane. $B_{0}=\;\frac{\mu _{0} m}{4 \pi } \;\frac{1}{r^{2}}$ $B_{H}=2 B_{0} \cos \theta$ $\Rightarrow \;\; 0.4 \times 10^{-4}=\;\frac{2 \mu _{0} m}{4 \pi r^{2}} \cdot \;\frac{7 \times 10^{-2}}{r}$ $( \because B_{H}=0.4 {G}=0.4 \times 10^{-4} {T})$ $[ \because \cos \theta =\;\frac{7}{r}$ and $r=(7^{2}+18^{2})^{1 {/} 2}]$ $\Rightarrow \;\; 0.4 \times 10^{-4}=2 \times 10^{-7} \times \;{m \times 7}/{(7^{2}+18^{2})^{3 {/} 2}} \times 10^{4}$ $\Rightarrow \;\; m=\;{4 \times 10^{-2} \times (373)^{3 {/} 2}}/{14}$...(i) $\because$ Magnetic moment, $M=m \times 2 l=m \times \;\frac{14}{100}$...(ii) From Eqs. (i) & (ii), we get $M \;\; =\;{4 \times 10^{-2} \times (373)^{3{/} 2}}/{14} \times \;\frac{14}{100}$ $\;\; =4 \times 10^{-4} \times 7203.82=2.88 \frac{J}{T}$

Since, in ferromagnetic material, with increase in temperature susceptibility decreases, $\therefore$ Ferromagnetic material below Curie temperature will show saturation magnetisation. Hence, option (b) is the correct i.e. domain is defined as a macroscopic region with saturation magnetisation.

${B}=\mu\.({H}+{I})$ ${B}=\mu\.{H}(1+\frac{{I}}{{H}})$ ${B}={B}_{0}(1+{x})$ ${B}-{B}_{0}={B}_{0} {x}$ $\frac{{B}-{B}_{0}}{{B}_{0}}={x}$ $\frac{{B}-{B}_{0}}{{B}_{0}} \\times 100=100 {x}$ $=2.2 \\times 10^{-3}=\frac{22}{10^{4}}$

A perfect diamagnetic substance will completely expel the magnetic field. Therefore, there will be no magnetic field inside the cavity of sphere. Hence the paramagnetic substance kept inside the cavity will experience no force.

For paramagnetic material According to curies law $\chi \propto\frac{1}{\ {T}}$ $\chi \propto\frac{1}{ {T}}\Rightarrow \chi _{1} {T}_{1}=\chi _{2} {T}_{2}$ $\Rightarrow \frac{6}{0.4} \\times 4=\frac{{I}}{0.3} \\times 24$ $\ {I}= \frac{0.3}{0.4}=0.75 {A} / {m}$

Torque on a bar magnet $: {I}= {MB} \sin \theta$ Here, $\theta =3o°, {I}=0.018 {N}- {m}, {B}=0.06 {T}$ $\Rightarrow0.018= {M} \\times 0.06 \\times \sin 30°$ $\Rightarrow 0.018= {M} \\times 0.06 \\times \frac{1}{2}$ $\Rightarrow {M}=0.6 {A}- {m}^{2}$ Now ${v}=- {MB} \cos \theta$ Position of stable equilibrium $(\theta =0°):$ ${u}_{ {i}}=- {MB}$ Position of unstable equilibrium $(\theta =180°):$ ${u}_{ {f}}= {MB}$ $\Rightarrow$ work done $: \Delta{U}$ $\Rightarrow {W}=2 \ {MB}$ $\Rightarrow{W}=2 \\times 0.6 \\times 0.06$ $\Rightarrow {W}=7.2 \\times 10^{-2} \ {J}$ option (4) is correct

As for permanent magnet large retentivity and large coercivity required

Retentivity = 1.0 TCo-ercivity = 50 A/mSaturation = 1.5 T

Work done , W = $(\Delta \mu ^{\to }).\vec{B}$ = 2 × $10^{-2}$ × 1 cos (0.125) = 0.02 J ∴ correct answer is (2)

µBsin 45° = F $l/2$ sin 45° F = 2µB

x α $1/T_C$ curie law for paramagnetic substane $x_1/x_2$ = $T_{C_2}/T_{C_1}$ ${2.8\times 10^{-4}}/x_2$ = $\frac{300}{350}$ $x_2$ = $\frac{2.8\times 350\times 10^{-4}}{300}$ = 3.266 × $10^{-4}$

$B_{1}= \frac{2 \mu_{0} M}{4 \pi (d / 2)^{3}}$ $B_{2}= \frac{2 \mu_{0} M}{4 \pi (d / 2)^{3}}$Force $=q( {V} ^{\to }\times \vec{B})=0$ $(\because\vec{V} \| \vec{B})$

Coercivity = H = $B/\mu _0$ = ni = $N/l$ i = $\frac{100}{0.2}$ × 5.2 = 2600 A/m

X = $I/H$ I = $\frac{\text{Magnetic} \text{moment}}{\text{Volume}}$ I = $\frac{20\times 10^{-6}}{10^{-6}}$ = 20 N/$m^2$ X = $\frac{20}{60\times 10^{+3}}$ = $1/3 \times 10^{-3}$ = 0.33 × $10^{-3}$ = 3.3 × $10^{-4}$

$n= \frac{1}{2 \pi } \sqrt{ \frac{B \cos \theta }{I}}$ $\frac{n_{1}}{n_{2}}=\sqrt{ \frac{B_{1} \cos \theta _{1}}{B_{2} \cos \theta _{2}}}=30/40$ ${B_{1} \times \frac{1}{\sqrt{2}}}/{B_{2}\times \frac{\sqrt{3}}{2}}= \frac{9}{16}$ $\frac{B_{1}}{B_{2}}= \frac{9}{16} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{1}=0.7$