Physics Ncert Solutions Class 12th

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Let us take a point p to be at a distance z from the center of ring

V=K $\int \frac{dq}{r}=k$ $\frac{dq}{\sqrt[]{z^2+a^2}}$  = $\frac{k}{\sqrt[]{z^2+a^2}}\int dq$ = $\frac{kQ}{\sqrt[]{z^2+a^2}}$

 

U= w =qV= $\frac{kQ(-q)}{\sqrt[]{z^2+a^2}}$ = $\frac{-KQq}{\sqrt[]{z^2+a^2}}$

Charge would perform oscillations.

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According to the relation that E= - $\frac{dv}{dr}$ , potential always decrease in the direction of electric field . so if we go from any point to infinity that decreases the potential.

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If there is a potential then E? 0 electric field comes into existence, which is given by as E=-dV/dr
It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. Hence, the entire volume inside must be equipotential.

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As the electric field and force both are conservative so work done in conservative body as the body is closed is zero.

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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.

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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.

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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.

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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