- Home
- Standard 12
- Physics
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2\, sec$ in the 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....$s$
$1$
$2$
$3$
$4$
Solution
The time period $T$ of oscillation of a magnet is given by:
$T = 2\pi \sqrt {\frac{1}{{MB}}} $
where,
$\mathrm{I}=$ Moment of inertia of the magnet about the axis of rotation
$\mathrm{M}=$ Magnetic moment of the magnet
$\mathrm{B}=$ Uniform magnetic field
As the $I$, $B$ remains the same
$\therefore $ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}}} \text { or } \frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\sqrt{\frac{\mathrm{B}_{1}}{\mathrm{B}_{2}}}$
According to given problem.
${\mathrm{B}_{1}=24 \,\mu \mathrm{T}}$
${\mathrm{B}_{2}=24\, \mu \mathrm{T}-18\, \mu \mathrm{T}=6\, \mu \mathrm{T}}$
${\mathrm{T}_{1}=2\, \mathrm{s}}$
$\therefore $ ${\mathrm{T}_{2}=(2\, \mathrm{s}) \sqrt{\frac{(24\, \mu \mathrm{T})}{(6\, \mu \mathrm{T})}}=4 \,\mathrm{s}}$
