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Permanent Magnets — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsPermanent Magnets4 Marks
Question
A short magnet makes $40$ oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is $2\mu\text{T.}$ Another short magnet of magnetic moment $1.6A-m^2$ is placed $20\ cm$ east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole:
  1. Towards north.
  2. Towards south.
✓

Answer

$\gamma_1=40$ oscillations/ minute
$\text{B}_\text{H}=25\mu\text{T}$
$m$ of second magnet $= 1.6A-m^2$
$d = 20\ cm = 0.2m$
For north facing north
$\gamma_1=\frac{1}{2\pi}\sqrt{\frac{\text{MB}_{\text{H}}}{\text{I}}}$
$\gamma_2=\frac{1}{2\pi}\sqrt{\frac{\text{M}(\text{B}_{\text{H}}-\text{B})}{\text{I}}}$
$\text{B}=\frac{\mu_0}{4\pi}\frac{\text{m}}{\text{d}^3}=\frac{10^{-7}\times1.6}{8\times10^{-3}}=20\mu\text{T}$
$\frac{\gamma_1}{\gamma_2}=\sqrt{\frac{\text{B}}{\text{B}_{\text{H}-\text{B}}}}$
$\Rightarrow\frac{40}{\gamma_2}=\sqrt{\frac{25}{5}}$
$\Rightarrow\gamma_2=\frac{40}{\sqrt{5}}$
$=17.88\approx18\text{ osci/min}$ 
For north pole facing south
$\gamma_1=\frac{1}{2\pi}\sqrt{\frac{\text{MB}_{\text{H}}}{\text{I}}}$
$\gamma_2=\frac{1}{2\pi}\sqrt{\frac{\text{M}(\text{B}_{\text{H}}-\text{B})}{\text{I}}}$
$\frac{\gamma_1}{\gamma_2}=\sqrt{\frac{\text{B}}{\text{B}_{\text{H}-\text{B}}}}$
$\Rightarrow\frac{40}{\gamma_2}=\sqrt{\frac{25}{45}}$
$\Rightarrow\gamma_2=\frac{40}{\sqrt{\Big(\frac{25}{45}\Big)}}$
$=53.66\approx54\text{ osci/min}$

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