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

$$\eqalign{ & F = mB\,\,{\text{or}}\,\,6 \times {10^{ - 4}} = m \times 2 \times {10^{ - 5}} \cr & \therefore m = 30\,A - m \cr & {\text{Now,}}\,\,\ell = \frac{M}{m} = \frac{3}{{30}} = 0.1\,m \cr} $$

We know that $$\frac{B}{{{B_H}}} = \tan \theta $$
$$\eqalign{ & {\text{or}}\,B = {B_H}\tan \theta \cr & = 0.34 \times {10^{ - 4}}\tan {30^ \circ } \cr & = 1.96 \times {10^{ - 5}}T \cr} $$

As magnetic susceptibility $${\chi _m} \propto \frac{1}{T},$$   therefore
$$\eqalign{ & \frac{{{\chi _2}}}{{{\chi _1}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{\chi _2}}}{{0.0060}} = \frac{{273 - 73}}{{273 - 173}} = \frac{{200}}{{100}} = 2 \cr & {\chi _2} = 2 \times 0.0060 = 0.0120 \cr} $$

Initially magnetic moment of system
$${M_1} = \sqrt {{M^2} + {M^2}} = \sqrt {2M} $$
and moment of inertia
$${I_1} = I + I = 2I.$$
Finally when one of the magnet is removed then
$$\eqalign{ & {M_2} = M\,{\text{and}}\,{I_2} = I \cr & {\text{So,}}\,T = 2\pi \sqrt {\frac{I}{{M{B_H}}}} \cr & \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}} \times \frac{{{M_2}}}{{{M_1}}}} = \sqrt {\frac{{2I}}{I} \times \frac{M}{{\sqrt 2 M}}} \cr & \Rightarrow {T_2} = \frac{{{2^{\frac{5}{4}}}}}{{{2^{\frac{1}{4}}}}} = 2\,\sec \cr} $$

Magnetism of a magnet falls with rise of temperature and becomes practically zero above curie temperature.

Let the $${B_H}$$ and $${B_V}$$ be the horizontal and vertical component of earth’s magnetic field $$B.$$

$$\tan \theta = \frac{{{B_V}}}{{{B_H}}} \Rightarrow \cot \theta = \frac{{{B_H}}}{{{B_V}}}\,......\left( {\text{i}} \right)$$
Let, plane 1 and 2 are mutually perpendicular planes making angle $$\theta $$ and $$\left( {{{90}^ \circ } - \theta } \right)$$  with magnetic meridian. The vertical component of earth of earth's magnetic field remain same in two plane but effective horizontal components in the two planes is given by
$$\eqalign{ & {B_1} = {B_H}\cos \theta '\,\,......\left( {{\text{ii}}} \right) \cr & {\text{and}}\,\,{B_2} = {B_H}\sin \theta '\,\,......\left( {{\text{iii}}} \right) \cr & {\text{Then,}}\,\,\tan {\theta _1} = \frac{{{B_V}}}{{{B_1}}} = \frac{{{B_V}}}{{{B_H}\cos \theta '}} \cr & \cot {\theta _1} = \frac{{{B_H}\cos \theta '}}{{{B_V}}}\,......\left( {{\text{iv}}} \right) \cr} $$
Similarly,
$$\eqalign{ & \Rightarrow \tan {\theta _2} = \frac{{{B_V}}}{{{B_2}}} = \frac{{{B_V}}}{{{B_H}\sin \theta '}} \cr & \Rightarrow \cot {\theta _2} = \frac{{{B_H}\sin \theta '}}{{{B_V}}}\,......\left( {\text{v}} \right) \cr} $$
From Eq. (iv) and Eq. (v)
$$\eqalign{ & \Rightarrow {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2} = \frac{{B_H^2{{\cos }^2}\theta '}}{{B_V^2}} + \frac{{B_H^2{{\sin }^2}\theta '}}{{B_V^2}} \cr & \Rightarrow {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2} = \frac{{B_H^2}}{{B_V^2}}\left( {{{\cos }^2}\theta ' + {{\sin }^2}\theta '} \right) \cr & \Rightarrow {\cot ^2}{\theta _1} + {\cot ^2}{\theta _2} = {\cot ^2}\theta \cr} $$

We know that,
$$\eqalign{ & {\mu _r} = 1 + \chi \cr & {\text{or}}\,\,\chi = {\mu _{r - 1}} = 5500 - 1 = 5499 \cr} $$

$$\eqalign{ & {U_i} = - MB\cos {0^ \circ } = - MB \cr & {\text{and}}\,{U_f} = - MB\cos {360^ \circ } = - MB \cr & {\text{Now,}}\,W = {U_f} - {U_i} = - MB - \left( { - MB} \right) = 0. \cr} $$

Magnetic moment, $$M = 8.7 \times {10^{ - 2}}A{m^2}$$
moment of inertia, $$I = 11.5 \times {10^{ - 6}}kg\,{m^2}$$     Time period of oscillation is
$$\eqalign{ & T = \frac{{6.70}}{{10}} = 0.6755\,{\text{As,}}\,T = 2\pi \sqrt {\frac{I}{{MB}}} ;B = \frac{{4{\pi ^2}{I^2}}}{{M{T^2}}} \cr & \therefore B = \frac{{4 \times {{\left( {3.14} \right)}^2} \times 11.5 \times {{10}^{ - 6}}}}{{8.7 \times {{10}^{ - 2}} \times {{\left( {0.67} \right)}^2}}} = 0.012\,T \cr} $$

KEY CONCEPT : The temperature above which a ferromagnetic substance becomes paramagnetic is called Curie’s temperature.