Physics NCERT Exemplar Solutions Class 12th Chapter Five

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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

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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

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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

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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

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

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

$3R=a$

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Explanation- P is also on the magnetic equator, so the angle of dip= 0, because the value of angle of Dip at equator is zero. Q is also on the magnetic equator, thus the angle of dip is zero.

As earth tilted on its axis by 11.3°, thus the declination at Q is11.3°.

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Explanation- as  is dimensionless quantity , it should have no involvement of charge Qin its dimensional formula

Let $\chi ={\mu }_0{e}^2{m}^a{v}^b{R}^c$

[ $M^0{L}^0{T}^0{Q}^0$ ]= [ML $Q^{-2}$ ] $*\left[Q^2\right]\left[M^a\right]*{\left[L{T}^{-1}\right]}^b{\left[L\right]}^c$

   = [ $M^{1+a}+L^{1+b+c}{T}^{-b}{Q}^0$ ]

After solving we get a=-1 , b= 0, c =-1

Putting these values in equations

$\chi ={\mu }_o{e}^2{m}^{-1}{v}^2{R}^{-1}=\frac{{\mu }_0{e}^2}{mR}$

By using their standard values we get $\chi$ =10 which proves it is dimensionless quantity

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Explanation- the magnetic field induction at a pojnt z distance from dipole moment is

B= $\frac{{\mu }_o}{4\pi }\frac{2M}{z^3}$

Along z axis from p to q

${\int }_p^{Q}B.dl={\int }_P^{Q}Bdlcos0$ = ${\int }_P^{Q}Bdz$

= ${\int }_a^R\frac{{\mu }_o}{2\pi }\frac{M}{z^3}dz=\frac{{\mu }_{o}M}{2\pi }\left(-\frac{1}{2}\right)\left(\frac{1}{R^2}-\frac{1}{a^2}\right)$

= $\frac{{\mu }_{o}M}{4\pi }\left(\frac{1}{a^2}-\frac{1}{R^2}\right)$

(b) the point A lies on the equatorial line of the magnetic dipole of moment Msin $\theta$

B= $\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{R^3}$ dl= Rd $\theta$

$\int B.dl=\int Bdlcos\theta ={\int }_0^{\frac{\pi }{2}}\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{R^3}Rd\theta$

Circular arc = $\frac{{\mu }_o}{4\pi }\frac{M}{R}(-cos\theta )$ = $\frac{{\mu }_o}{4\pi }\frac{M}{R^2}$

(c) along x axis over the path ST

From figure every point lies on the equatorial line of magnetic dipole so

B= $\frac{{\mu }_o}{4\pi }\frac{M}{x^3}$   the value of B is zero as the angle between M and dl is 90 

(d)along the quarter circle TP of radius a

B along circular arc $\int B.dl={\int }_{\pi /2}^0\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{a^3}$ d $\theta$

B = $-\frac{{\mu }_o}{4\pi }\frac{M}{a^2}$

Net magnetic field through PQST  is zero