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

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

As  we know $\aleph \propto$ $\rho$

$\frac{{\rho N}_2}{\rho Cu}=$  = $\frac{{\aleph N}_2}{\aleph Cu}$ 1.6 $\times$ 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 =

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°

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&nb

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 $Q^{-2}$ ] $\times \left[Q^2\right]\left[M^a\right]\times {\left[L{T}^{-1}\right]}^b{\left[L\right]}^c$

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

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

Putting these values in equation

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