A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $$2\,s$$  in the earth’s horizontal magnetic field of $$24\,\mu T.$$  When a horizontal field of $$18\,\mu T$$  is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be

Solution :
Problem Solving Strategy
Find out the relation of time period with the earth’s horizontal magnetic field and then compare it for the two given cases.
Time period in vibration magnetometer is given by
$$\eqalign{ & T = 2\pi \sqrt {\frac{I}{{M \times {B_H}}}} \cr & \Rightarrow T \propto \frac{1}{{\sqrt {{B_H}} }} \cr & {\text{So, for two different cases}} \cr & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{{\left( {{B_H}} \right)}_2}}}{{{{\left( {{B_H}} \right)}_1}}}} \cr & \Rightarrow \frac{2}{{{T_2}}} = \sqrt {\frac{{18}}{{24}}} \cr & \therefore T = 2.3\,s \approx 2\,s \cr} $$