Q124 / 200 · NEET 2010 (Screening) · Physics
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \mathrm{~s}$ in earth's horizontal magnetic field of $24 \mu \mathrm{T}$. When a horizontal field of $18 \mu \mathrm{T}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
- A$1 \mathrm{~s}$
- B$2 \mathrm{~s}$
- C$3 \mathrm{~s}$
- Correct answerD$4 \mathrm{~s}$
Solution
The time period T of oscillation of a magnet is given by
\(T=2 \pi \sqrt{\frac{I}{M B}}\)
where,
I = Moment of inertia of the magnet about the axis of rotation
\(M=\) Magnetic moment of the magnet
\(B=\) Uniform magnetic field As I, B remains the same
\(\therefore \quad T \propto \frac{1}{\sqrt{B}}\) or \(\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}\)
According to given problem,
\(\begin{aligned}
B_1 & =24 \mu \mathrm{T} \\
B_2 & =24 \mu \mathrm{T}-18 \mu \mathrm{T}=6 \mu \mathrm{T} \\
T_1 & =2 \mathrm{~s} \\
\therefore \quad T_2 & =(2 \mathrm{~s}) \sqrt{\frac{(24 \mu \mathrm{T})}{(6 \mu \mathrm{T})}}=4 \mathrm{~s}
\end{aligned}\)
Hence, the correct answer is option: 'D'.