Question

A magnet is placed in the magnetic meridian with its north pole pointing north. The neutral point is obtained at a point $\text{P}$ towards east of the middle point of the magnet. When the magnet is reversed end to end at the same place, a magnetic needle completes $12$ oscillations per $\text{minute}$ at the point $\text{P}$. If the magnet be removed, how many oscillations will the needle perform per $\text{minute}$ ?

• A. $6\sqrt{2}$
• B. $6$
• C. $12$
• D. $12\sqrt{2}$

Answer 1

The correct option is A $6\sqrt{2}$

Given:
$f_1=12{\text{min}}^{-1}$

We know that, at the neutral point,

$B_{magnet}=B_H$

$\text{Case I:}$ When the bar magnet is reversed end to end position, then the net magnetic field at $\text{P}$ be,

$B_{\text{I}}=B_{magnet}+B_H=2B_H$


$\text{Case II:}$When the short magnet is removed, the net magnetic field at point $\text{P}$ will be equal to the Earth's horizontal magnetic field at this point. So,

$B_{\text{II}}=B_H$

We know that the frequency of oscillation is given by,

$f=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$

So, for $\text{case I}:$

$\frac{12}{60}=\frac{1}{2\pi }\sqrt{\frac{M\left(2B_H\right)}{I}}...\left(1\right)$

For $\text{case II}:$

$f_2=\frac{1}{2\pi }\sqrt{\frac{M{B}_H}{I}}...\left(2\right)$

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

$f_2=\frac{12/60}{\sqrt{2}}=\frac{12}{60\sqrt{2}}\text{Hz}$

$\therefore f_2=\frac{12}{60\sqrt{2}}\times 60=6\sqrt{2}{\text{min}}^{-1}$

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