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Permanent Magnets — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsPermanent Magnets4 Marks
Question
A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by this combination. Neglect any induced magnetism.
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Answer

$\gamma_1=40$ oscillations/minute

$\text{B}=25\mu\text{T}$

M of second magnet $=1.6\text{A-m}^2$

$\text{d}=20\text{cm}=0.2\text{m}$

  1. For north facing north.

$\gamma=\frac{1}{2\pi}\sqrt{\frac{\text{MB}_\text{H}}{\text{I}}}$

$\gamma=\frac{1}{2\pi}\sqrt{\frac{\text{M}(\text{B}_\text{H}-\text{B})}{\text{I}}}$

$\text{B}=\frac{\mu_0}{4\pi}\frac{\text{m}}{\text{d}^3}=\frac{10^{-7}\times1.6}{8\times10^{-3}}=20\mu\text{T}$

$\frac{\gamma_1}{\gamma_2}=\sqrt{\frac{\text{B}}{\text{B}_\text{H}-\text{B}}}\Rightarrow\frac{40}{\gamma _2}=\sqrt{\frac{25}{5}}$

$\Rightarrow\gamma_2=\frac{40}{\sqrt{5}}=17.88\approx \text{osci/min}$

  1. 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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