Magnetic Materials MCQ Questions & Answers in Electrostatics and Magnetism | Physics

When magnetic dipole is rotated from initial position $${\theta _1}$$ to final position $${\theta _2},$$
then work done $$ = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$$
Given, $${\theta _1} = {0^ \circ },{\theta _2} = {60^ \circ }$$
Magnetic moment, $$M = 2 \times {10^4}\,J/T$$
Magnetic field, $$B = 6 \times {10^{ - 4}}T$$
So, $$W = MB\left( {1 - \frac{1}{2}} \right)\,\,\left[ {\because \cos {0^ \circ } = 1\,{\text{and}}\,\cos {{60}^ \circ } = \frac{1}{2}} \right]$$
$$ = \frac{{2 \times {{10}^4} \times 6 \times {{10}^{ - 4}}}}{2} = 6\,J$$

When a diamagnetic material is placed in an external magnetic field, the spin motion of electrons is so modified that the electrons which produce the magnetic moments in the direction of external field slow down while the electrons which produce magnetic moments in opposite direction get accelerated.
Thus, a net magnetic moment is induced in the opposite direction of applied magnetic field. Hence, the substance is magnetised opposite to the external field. Thus, it moves from stronger to weaker parts of the magnetic field.
Diamagnetism is present in all the substances. However, its effect is so weak in most cases that it gets shifted by other effects like para-magnetism, ferromagnetism etc.

The magnetic susceptibility of a material is a measure of the ease with which a specimen of that material can be magnetised in a magnetising field. For a diamagnetic substance, magnetic susceptibility $$\left( {{\chi _m}} \right)$$  is independent of temperature.

Magnetic field in solenoid $$B = {\mu _0}ni$$
$$\eqalign{ & \Rightarrow \frac{B}{{{\mu _0}}} = ni\,\left( {{\text{where}}\,n = {\text{number}}\,{\text{of}}\,{\text{turns}}\,{\text{per}}\,{\text{unit}}\,{\text{length}}} \right) \cr & \Rightarrow \frac{B}{{{\mu _0}}} = \frac{{Ni}}{L} \cr & \Rightarrow 3 \times {10^3} = \frac{{100i}}{{10 \times {{10}^{ - 2}}}} \cr & \Rightarrow i = 3A \cr} $$

Magnetic moment, $$M = m\ell $$
$$\frac{M}{\ell } = m,$$   where $$m$$ is the pole strength.
Therefore distance between poles
$$\eqalign{ & = \sqrt {{{\left( {\frac{\ell }{2}} \right)}^2} + {{\left( {\frac{\ell }{2}} \right)}^2}} \cr & = \frac{\ell }{{\sqrt 2 }} \cr & {\text{So,}}\,M' = \frac{{m\ell }}{{\sqrt 2 }} = \frac{M}{{\sqrt 2 }} \cr} $$

$$\frac{{\tan {\theta _1}}}{{\tan {\theta _2}}} = \frac{2}{1}$$

$${\text{Here,}}\,\,2\ell = 8\,cm,\ell = 4\,cm,\;d = \frac{6}{2} = 3\,cm.$$
At neutral point,
$$\eqalign{ & H = B = \frac{{{\mu _0}}}{{4\pi }}\frac{M}{{{{\left( {{d^2} + {\ell ^2}} \right)}^{\frac{3}{2}}}}} \cr & = {10^{ - 7}}\frac{M}{{{{\left( {5 \times {{10}^{ - 2}}} \right)}^3}}} = \frac{M}{{1250}} \cr & \therefore M = 1250\,H = 1250 \times 3.2 \times {10^{ - 5}}A{m^2} \cr & m = \frac{M}{{2\ell }} = \frac{{1250 \times 3.2 \times {{10}^{ - 5}}}}{{8 \times {{10}^{ - 2}}}}Am. = 0.5\,Am = 0.5 \times \frac{1}{{10}}ab\,amp \times 100\,cm \cr & = 5\,ab{\text{-}}amp\,cm. \cr} $$

$$\eqalign{ & i = \left( {\frac{{2R{B_H}}}{{{\mu _0}N}}} \right)\tan \theta \cr & {\text{or}}\,\frac{V}{R} = \left( {\frac{{2R{B_H}}}{{{\mu _0}N}}} \right)\tan \theta \,......\left( {\text{i}} \right) \cr} $$
When number of turns are doubled, resistance of the coil is also doubled, so
$$\frac{V}{{\left( {2R} \right)}} = \left[ {\frac{{2R{B_H}}}{{{\mu _0}\left( {2N} \right)}}} \right]\tan \theta '\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii),
$$\theta ' = \theta .$$

For a diamagnetic material, the value of $${\mu _r}$$ is less than one. For any material, the value of $${ \in _r}$$ is always greater than 1.

$$\eqalign{ & \frac{{\tan {\theta _2}}}{{\tan {\theta _1}}} = \frac{{d_1^3}}{{\;d_2^3}} = \frac{{{r^3}}}{{{{\left[ {r{{\left( 3 \right)}^{\frac{1}{3}}}} \right]}^3}}} = \frac{1}{3} \cr & \tan {\theta _2} = \frac{1}{3}\tan {\theta _1} = \frac{{\tan 60}}{3} = \frac{{\sqrt 3 }}{3} = \frac{1}{{\sqrt 3 }} \cr & \therefore {\theta _2} = {30^ \circ } \cr} $$