Question

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

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Answer 1

Answer: (D)

The time period $T$ of oscillation of a magnet is given by
$T=2\pi \sqrt{\frac{1}{MB}}$
where,
$I=$ Moment of inertia of the magnet about the axis of rotation
$M=$ Magnetic moment of the magnet
$B=$ Uniform magnetic field
As the $I,B$ remains the same
$\therefore T\propto \frac{1}{\sqrt{B}}$
or $\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}$
According to given problem,
$B_1=24\mu T$
$B_2=24\mu T-18\mu T=6\mu T$
$T_1=2s$
$\therefore T_2=(2s)=\sqrt{\frac{(24\mu T)}{(6\mu T)}}=4s$