Question

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where the dip angle is 30 degrees and 15 oscillations per minute at a place where the dip angle is 60 degrees. The ratio of the total earth’s magnetic field at the two places is.
A. $3\sqrt{3}\colon8$
B. $16\colon9\sqrt{3}$
C. $2\sqrt{3}\colon9$
D. $4\colon9$

Answer 1

Hint: Determine the frequency for both the oscillations. Also, frequency is the reciprocal of time period. Use the relation of time period in frequency formula.Rearrange the equation in the form of the ratio of magnetic field. By doing the necessary substitution, determine the ratio of the total earth’s magnetic field at the two places.

Formula used:
$\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{MB_{H}}{I}}$
Where, B is the magnetic field and I is the moment of inertia.
 $\nu=\dfrac{1}{T}$
Where, T is the time period and $\nu$ is the frequency.

Complete step by step solution:
We know, frequency is the no. of oscillations per minute,
Thus, $\nu_{1}=\dfrac{20}{60}$
$\Rightarrow\nu_{1}=\dfrac{1}{3} sec^{-1}$
Similarly,
$\nu_{2}=\dfrac{15}{60}$
$\Rightarrow\nu_{2}=\dfrac{1}{4} sec^{-1}$
Also, frequency is the reciprocal of time period. Therefore,
$\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{MB_{H}}{I}}$
This implies,
$\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{MB\cos\Phi}{I}}$
Where, $B_{H}=B\cos\Phi$
Now, we can write
$\dfrac{\nu_{1}}{\nu_{2}}=\sqrt{\dfrac{B_{1}\cos\Phi_{1}}{B_{2}\cos\Phi_{2}}}$
Squaring both sides, we get
$\lgroup\dfrac{\nu_{1}}{\nu_{2}}\rgroup^{2}=\dfrac{B_{1}\cos\Phi_{1}}{B_{2}\cos\Phi_{2}}$
$\Rightarrow\dfrac{B_{1}}{B_{2}}=\lgroup\dfrac{\nu_{1}}{\nu_{2}}\rgroup^{2}\lgroup\dfrac{cos\phi_{2}}{cos\phi_{1}}\rgroup$ ……(i)
Here, $\phi_{1}=30$ and $\phi_{2}=60$
Substituting all the values in equation (i), we get
$\dfrac{B_{1}}{B_{2}}=\lgroup\dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}\rgroup^{2}\lgroup\dfrac{\cos60}{\cos30}\rgroup$
This implies,
$\dfrac{B_{1}}{B_{2}}=\dfrac{16}{9}\times\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$
$\dfrac{B_{1}}{B_{2}}==\dfrac{16}{9\sqrt{3}}$

Hence, the correct option is B.

Note: Many make the mistake of using the time period formula of simple pendulum. Here we need to use the relation related to the magnetic field. Be careful while rearranging the equation for the ratio of the magnetic field.