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
$T=2 \pi \sqrt{\frac{1}{M B}}$
where,
$I=$ Moment of inertia of the magnet about the axis of rotation
$M=$ Magnetic moment of the magnet
$B=$ Uniform magnetic field
As 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}=2 \,s $
$\therefore T_{2}=(2 s)=\sqrt{\frac{(24 \mu T)}{(6 \mu T)}}=4 \,s$