Question

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2\sec$ in the earth's horizontal magnetic field of $24$ microteslas. When a horizontal field of $18$ microteslas is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be

• A. $1\mathrm{s}$
• B. $2\mathrm{s}$
• C. $3\mathrm{s}$
• D. $4\mathrm{s}$

Answer 1

The correct option is D

$4\mathrm{s}$

Step 1: Given data

The magnet executes oscillations with a time period of $2\sec$.

The earth's horizontal magnetic field of $24$ microteslas.

A horizontal field of $18$ microteslas.

Step 2: To find

We have to determine the new time period of the magnet.

Step 3: Calculation

We know that,

The time period T of oscillation of a magnet is:

$\Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$

Here,

l is the moment of inertia of the magnet about the axis of rotation.

M is the magnetic moment of the magnet.

B is the uniform magnetic field as l, remains the same.

Now,

$\mathrm{T}\propto \frac{1}{\sqrt{\mathrm{B}}}$ or, $\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}=\sqrt{\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}}$

Then,

$$\begin{gathered} {\mathrm{B}}_1=24\mathrm{ut} \\ {\mathrm{B}}_2=24\mathrm{ut}–18\mathrm{ut} \\ =6\mathrm{ut} \\ {\mathrm{T}}_1=2\mathrm{seconds} \\ {\mathrm{T}}_2=2\sqrt{\frac{24\mathrm{ut}}{6\mathrm{ut}}} \\ =2\sqrt{4} \\ =2\times 2 \\ =4\mathrm{s} \end{gathered}$$

Therefore, option D is the correct choice.