Question

The frequency of oscillation of a bar magnet in an oscillation magnetometer in the earth's magnetic field is $40$ oscillations per $\text{minute}$. A short bar magnet is placed to the north of the magnetometer, at a separation of $20\text{cm}$ from the oscillating magnet, with its north pole pointing towards north as shown in figure. The frequency of oscillation is found to be increases to $60$ oscillations per $\text{minute}$. Calculate the magnetic moment of this short bar magnet.
[Horizontal component of the earth's magnetic field is $24\mu \text{T}$]

• A. $1.2{\text{Am}}^2$
• B. $12{\text{Am}}^2$
• C. $0.12{\text{Am}}^2$
• D. $120{\text{Am}}^2$

Answer 1

The correct option is A $1.2{\text{Am}}^2$

$\text{Case I:}$
In the presence of only earth's magnetic field. The frequency of oscillation of the magnet in the magnetometer is,

$f=\frac{1}{2\pi }\sqrt{\frac{M{B}_H}{I}}.....\left(1\right)$

Where, $B_H$ is the horizontal component of earth magnetic field.

$\text{Case II:}$
When a short magnet is placed at a distance of $20\text{cm}$ from the magnetometer.


The net magnetic field on the magnetometer is,

$\therefore B^{\prime }=B_{magnet}+B_H$

Now, the frequency of oscillation of the magnet is,

$f^{{}^{\prime }}=\frac{1}{2\pi }\sqrt{\frac{M{B}^{{}^{\prime }}}{I}}=\sqrt{\frac{M(B_{magnet}+B_H)}{I}}......\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{f^{{}^{\prime }}}{f}=\sqrt{\frac{B_{magnet}+B_H}{B_H}}$

$\Rightarrow {\left(\frac{f^{\prime }}{f}\right)}^2=\frac{B_{magnet}+B_H}{B_H}$

Given,

$f=40{\text{min}}^{-1}$ and $f^{{}^{\prime }}=60{\text{min}}^{-1}$

$\Rightarrow {\left(\frac{60}{40}\right)}^2=\frac{B_{magnet}+24\times {10}^{-6}}{24\times {10}^{-6}}$

$\Rightarrow B_{magnet}=30\times {10}^{-6}\text{T}=30\mu \text{T}$

If $M^{\prime }$ is the magnetic moment of the short magnet, then the magnetic field along its axis at a distance of $20\text{cm}$

$B_{magnet}=\frac{{\mu }_0}{4\pi }\left(\frac{2M^{{}^{\prime }}}{r^3}\right)$

$\Rightarrow 30\times {10}^{-6}=\frac{{10}^{-7}\times 2\times M^{\prime }}{(0.2{)}^3}$

$\Rightarrow M^{\prime }=\frac{30\times 8\times {10}^{-9}}{{10}^{-7}\times 2}$

$\Rightarrow M^{\prime }=1.2{\text{Am}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.