Magnetism and Matter

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

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

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

$3R=a$

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

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°.

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

Explanation- B_v= $\frac{{\mu }_o}{4\pi }\frac{2mcos\theta }{r^3}$

B_H=  $\frac{{\mu }_o}{4\pi }\frac{msin\theta }{r^3}$ adding and squaring both equations we get

B_v²+B_H²= $\frac{{\mu }_o}{4\pi }\frac{m^2}{r^6}[4{cos}^2\theta +{sin}^2\theta ]$

B= $\sqrt[]{B_v^2+B_v^2}=$ = $\frac{{\mu }_o}{4\pi }\frac{m}{r^3}[3{cos}^2\theta +1]$ ^1/2

The value of B is minimum if cos $\theta =\frac{\pi }{2}$

(b) tan $\delta =\frac{B_V}{B_H}=\frac{\mathrm{}\frac{{\mu }_o}{4\pi }\frac{2mcos\theta }{r^3}}{\frac{{\mu }_o}{4\pi }\frac{msin\theta }{r^3}\mathrm{ }}$ = 2cot $\theta$

For dip angle is zero

cot $\theta$  =0 and $\theta$ = $\frac{\pi }{2}$

(c) tan $\delta =\frac{B_V}{B_H}$ if angle of dip is 45

tan  $45=\frac{B_V}{B_H}$ so B_v=B_H

2cot $\theta$ =1

Cot $\theta$ =1/2 and tan $\theta$  = 2

 So $\theta =$ tan^-12

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

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

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

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