Class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- the system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- magnetic moment of dipole at origin o is

M=Mk

Magnetic field induction at p due to dipole moment

B= $\frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r

dS= r(rsin $\theta d\theta$ )r = (r^{2 $\theta$} sind $\theta r$ )

$\int B.ds$  = $\int \frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r(r²sin $\theta d\theta r$ )

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }2sin\theta cos\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }sin2\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}\left(\frac{-cos2\theta }{2}\right)$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ {cos4 $\pi -cos0$ }

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ (1-1)= 0

Explanation- When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.

(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation -  $\mathit{\aleph }$ = I/H

Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the susceptibility χ of diamagnetic material is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature is raised, the alignment is disturbed, resulting decrease in susceptibility of both with increase in temperature.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Answer- density of nitrogen = 28g/22.4L= 28g/22400cc

density of copper = 8g/22.4L= 8g/22400cc

on comparing $\frac{{\rho N}_2}{\rho Cu}$ = $\frac{28}{22400}*\frac{1}{8}$  = 1.6 $*$  10^-4

$\frac{{\aleph N}_2}{\aleph Cu}$ = $\frac{5*{10}^{-9}}{{10}^{-5}}$  = 5 $*$  10^-4

As  we know $\aleph \propto$ $\rho$

$\frac{{\rho N}_2}{\rho Cu}=$  = $\frac{{\aleph N}_2}{\aleph Cu}$ 1.6 $*$ 10^-4

This is a Short Answer Type Questions as classified in NCERT Exemplar

Answer- M (intensity of magnetisation) = 10⁶A/m

Length = 10cm = 10 $*$ 10^-2= 0.1m

M= I_m/l

I_m= M $*$ l= 10^{6 $*$} 0.1 = 10⁵A

This is a Short Answer Type Questions as classified in NCERT Exemplar

Answer – M= $\frac{eh}{4\pi m}$  or M $\propto$ 1/m

$\frac{M_p}{M_e}$ = $\frac{m_e}{m_p}$  = $\frac{M_e}{{1837M}_e}$ <<1

M_p<e

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know n = L/2 $\pi R$

Magnetic moment of circle = m₁= n₁IA₁= L.I $\pi R$ .²/2 $\pi$ R = LIR/2………(1)

Magnetic moment of square = m₂= n₂IA₂= $\frac{L}{4a}$ .I.a²= Lia/4…………….(2)

Moment of inertia of circle = MR²/2…………….(3)

Moment of inertia of square = Ma²/12…………….(4)

Frequency of circle f₁= 2 $\pi \sqrt[]{\frac{I_1}{m_{1}B}}$

Frequency of square f₂= 2 $\pi \sqrt[]{\frac{I_2}{m_{2}B}}$

As f₁=f₂

2 $\pi \sqrt[]{\frac{I_1}{m_{1}B}}$ =2 $\pi \sqrt[]{\frac{I_2}{m_{2}B}}$

So m₂/m₁= I₂/I₁

From eqn 1,2,3 and 4

$\frac{LIa.2}{4*LIR}$ = $\frac{M{a}^{2}2}{12M{R}^2}$

$\frac{a}{2R}=\frac{a^2}{6R^2}$

$3R=a$