Magnetic Materials MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Magnetic Materials MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
41.
If the period of oscillation of freely suspended bar magnet in earth’s horizontal field $$H$$ is $$4\,\sec.$$ When another magnet is brought near it, the period of oscillation is reduced to $$2s.$$ The magnetic field of second bar magnet is
42.
A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of $$30\,cm$$ from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $$A{m^2}$$ is close to: (Given $$\frac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}$$ in $$SI$$ units and $${B_H} = $$ Horizontal component of earth’s magnetic field $$ = 3.6 \times {10^{ - 5}}tesla$$ )
43.
A coil in the shape of an equilateral triangle of side $$l$$ is suspended between the pole pieces of a permanent magnet such that $$\overrightarrow B $$ is in plane of the coil. If due to a current $$i$$ in the triangle a torque $$\tau $$ acts on it, the side $$l$$ of the triangle is
A
$$\frac{2}{{\sqrt 3 }}{\left( {\frac{\tau }{{B.i}}} \right)^{\frac{1}{2}}}$$
B
$$2{\left( {\frac{\tau }{{\sqrt 3 B.i}}} \right)^{\frac{1}{2}}}$$
C
$$\frac{2}{{\sqrt 3 }}\left( {\frac{\tau }{{B.i}}} \right)$$
44.
A bar magnet has a length $$8\,cm.$$ The magnetic field at a point at a distance $$3\,cm$$ from the centre in the broad side-on position is found to be $$4 \times {10^{ - 6}}T.$$ The pole strength of the magnet is.
A
$$6 \times {10^{ - 5}}Am$$
B
$$5 \times {10^{ - 5}}Am$$
C
$$2 \times {10^{ - 4}}Am$$
D
$$3 \times {10^{ - 4}}Am$$
Answer :
$$6 \times {10^{ - 5}}Am$$
Magnetic field due to a bar magnet in the broad-side on position is given by
$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{M}{{{{\left[ {{r^2} + \frac{{{\ell ^2}}}{4}} \right]}^{\frac{3}{2}}}}};M = m\ell .$$
After substituting the values and simplifying we get
$$B = 6 \times {10^{ - 5}}A - m$$
45.
A $$30\,cm$$ long bar magnet is placed in the magnetic meridian with its north pole pointing south. The neutral point is obtained at a distance of $$40\,cm$$ from the center of the magnet. Find the pole strength of the magnet. The horizontal component of earth's magnetic field is $$0.34\,gauss.$$
A
$$25.7\,Am$$
B
$$23.7\,Am$$
C
$$28.7\,Am$$
D
$$26.7\,Am$$
Answer :
$$26.7\,Am$$
$$\eqalign{
& \therefore M = \frac{{4\pi }}{{{\mu _0}}}\frac{{{B_H}{{\left( {{r^2} - {\ell ^2}} \right)}^2}}}{{2r}} \cr
& = {10^7} \times \frac{{\left( {0.34 \times {{10}^{ - 4}}} \right).{{\left[ {{{\left( {0.40} \right)}^2} - {{\left( {0.15} \right)}^2}} \right]}^2}}}{{2 \times 0.40}} \cr
& = 8.0\,A{m^2} \cr} $$
The pole strength of the magnet is,
$$m = \frac{M}{{2\ell }} = \frac{{8.0}}{{0.30}} = 26.7\,Am$$
46.
The materials suitable for making electromagnets should have
A
high retentivity and low coercivity
B
low retentivity and low coercivity
C
high retentivity and high coercivity
D
low retentivity and high coercivity
Answer :
low retentivity and low coercivity
NOTE : Electro magnet should be amenable to magnetisation & demagnetization.
∴ retentivity should be low & coercivity should be low.
47.
According to Curie’s law, the magnetic susceptibility of a paramagnetic substance at an absolute temperature $$T$$ is proportional to
A
$$\frac{1}{{{T^2}}}$$
B
$${T^2}$$
C
$$\frac{1}{T}$$
D
$$T$$
Answer :
$$\frac{1}{T}$$
According to Curie's law, magnetic susceptibility $$\left( {{\chi _m}} \right)$$ of a paramagnetic substance is inversely proportional to absolute temperature $$\left( T \right)$$ i.e.
$${\chi _m} \propto \frac{1}{T}$$
48.
Let $$V$$ and $$H$$ be the vertical and horizontal components of earth's magnetic field at any point on earth. Near the north pole
A
$$V > > H$$
B
$$V < < H$$
C
$$V = H$$
D
$$V = H = 0$$
Answer :
$$V > > H$$
$$H = B\cos \theta ,V = B\sin \theta $$
Here $$B$$ = earth’s magnetic field
$$\eqalign{
& \theta = {\text{angle}}\,{\text{of}}\,{\text{dip}} = {90^ \circ }\,{\text{at}}\,{\text{north}}\,{\text{field}} \cr
& \Rightarrow H = B\cos {90^ \circ } = 0 \cr
& V = B\sin {90^ \circ } = B \cr
& \Rightarrow V > > H \cr} $$
49.
A bar magnet is oscillating in the earth’s magnetic field with a period $$T.$$ What happens to its period of motion, if its mass is quadrupled?
A
Motion remains simple harmonic with new period $$ = \frac{T}{2}$$
B
Motion remains simple harmonic with new period $$= 2 T$$
C
Motion remains simple harmonic with new period $$= 4 T$$
D
Motion remains simple harmonic and the period stays nearly constant
Answer :
Motion remains simple harmonic with new period $$= 2 T$$
The time period of a bar magnet in a magnetic field is given by
$$T = 2\pi \sqrt {\left( {\frac{I}{{MB}}} \right)} $$
where, $$I$$ is moment of inertia of bar magnet, $$M$$ is magnetic moment and $$B$$ is magnetic induction. When mass is made 4 times, moment of inertia $$I$$ becomes 4 times (as $$I = m{r^2},I \propto m$$ ). From the above equation of time period $$T \propto \sqrt I .$$ So, $$T$$ becomes twice as mass is quadrupled.
50.
If the susceptibility of dia, para and ferromagnetic materials are $${\chi _d},{\chi _p},{\chi _f}$$ respectively, then
A
$${\chi _d} < {\chi _p} < {\chi _f}$$
B
$${\chi _d} < {\chi _f} < {\chi _p}$$
C
$${\chi _f} < {\chi _d} < {\chi _p}$$
D
$${\chi _f} < {\chi _p} < {\chi _d}$$
Answer :
$${\chi _d} < {\chi _p} < {\chi _f}$$
$${\chi _d} < {\chi _p} < {\chi _f}$$
For diamagnetic substance $${\chi _d}$$ is small negative $$\left( {{{10}^{ - 5}}} \right)$$
For paramagnetic substances $${\chi _p}$$ is small and positive $$\left( {{{10}^{ - 3}}\,{\text{to}}\,{{10}^{ - 5}}} \right)$$
For ferromagnetic substances $${\chi _f}$$ is very large $$\left( {{{10}^{ 3}}\,{\text{to}}\,{{10}^{ 5}}} \right)$$