Magnetics Ques 102
- A paramagnetic substance in the form of a cube with sides 1 $cm$ has a magnetic dipole moment of $20 \times 10^{-6} J / T$ when a magnetic intensity of $60 \times 10^{3} A / m$ is applied. Its magnetic susceptibility is
(2019 Main, 11 Jan II)
(b) The atom is placed in a uniform magnetic induction $\mathbf{B}$ such that the normal to the plane of electron’s orbit makes an angle of $30^{\circ}$ with the magnetic induction. Find the torque experienced by the orbiting electron.
(a) $3.3 \times 10^{-4}$
(b) $3.3 \times 10^{-2}$
(c) $4.3 \times 10^{-2}$
(d) $2.3 \times 10^{-2}$
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Answer:
Correct Answer: 102.(a)
Solution:
- Given, side of cube $=1 cm=10^{-2} m$
$\therefore$ Volume, $V=10^{-6} m^{3}$
Dipole moment, $M=20 \times 10^{-6} J / T$
Applied magnetic intensity, $H=60 \times 10^{3} A / m$
Intensity of magnetisation
$$ I=\frac{M}{V}=\frac{20 \times 10^{-6}}{10^{-6}}=20 A / m $$
Now, magnetic susceptibility $\chi$ is
$\chi=\frac{\text { Intensity of magnetisation }}{\text { Applied magnetic intensity }}=\frac{I}{H}=\frac{20}{60 \times 10^{3}}$
$$ \Rightarrow \quad \chi=\frac{1}{3} \times 10^{-3}=3.33 \times 10^{-4} $$
BETA
