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

The Correct Option is 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,
$B_1 = 24\, \mu T$
$B_2 = 24\, \mu T - 18 \,\mu T = 6 \,\mu T$
$T_1 = 2\, s$
$\therefore T_2 = (2\, s) =\sqrt{\frac{(24\, \mu T)}{(6\, \mu T)}} = 4\, s$