Physics

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- for a charge particle to move in circular path

Mv²/r = qvB

v/r=w= bq/m

w= v/r = $\frac{L{T}^{-1}}{L}$ = T-1

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- force on charge particle due to both electric and magnetic field . now if electric force and magnetic field is zero then electric field and magnetic field is also zero.

Magnetic field is only responsible for constant velocity of charge particle if B=0 then E =0 if B is not equal to zero then electric field is also not equal to zero

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b, c, d)

Explanation- magnetic forces on both the particles cause equal acceleration because they both have equal velocity

Also they both gain or loose energy because direction of force in both cases is opposite.

And there is no change in center of mass of particles so it is determined by B alone.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b, c)

Explanation- if current is in opposite sense the? Bdl = 0, also there exist a point where b and dl is perpendicular.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b, d)

Explanation- F=IlB, the direction of force is given by fleming's left hand rule and F is perpendicular to the direction of magnetic field. So work done is zero.