Physics NCERT Exemplar Solutions Class 12th Chapter Five Magnetism And Matter

1. Verify the Ampere’s law for magnetic field of a point dipole of dipole moment m = mk . Take C as the closed curve running clockwise along .

(i) The z-axis from z = a > 0 to z = R;

(ii) Along the quarter circle of radius R and centre at the origin, in the first quadrant of x-z plane;

(iii) Along the x-axis from x = R to x = a, and (iv) along the quarter circle of radius a and centre at the origin in the first quadrant of x-z plane.

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

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 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

3. Assume the dipole model for earth’s magnetic field B which is given by

BV = vertical component of magnetic field = $\frac{{\mathit{\mu }}_{\mathit{o}}}{4\mathit{\pi }}\frac{2\mathit{m}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}{{\mathit{r}}^3}$

BH = Horizontal component of magnetic field = $\frac{{\mathit{\mu }}_{\mathit{o}}}{4\mathit{\pi }}\frac{\mathit{m}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{{\mathit{r}}^3}$ θ = 90° – lattitude as measured from magnetic equator. 

Find loci of points for which (i) |B| is minimum; (ii) dip angle is zero; and (iii) dip angle is ± 45°.

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

4. Consider the plane S formed by the dipole axis and the axis of earth. Let P be a point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q.

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°.