Question

A magnet makes 40 oscillation per minute at a place having magnetic intensity of $0.1\times {10}^{-5}$ Tesla. At another place it takes 2.5 sec to complete one oscillation. The value of earth's horizontal field at that place is

• A. $0.76\times {10}^{-6}T$
• B. $0.18\times {10}^{-6}T$
• C. $0.09\times {10}^{-6}T$
• D. $0.36\times {10}^{-6}T$

Answer 1

The correct option is D $0.36\times {10}^{-6}T$

Time period ofvibration ofa magnet is

$T=2\pi \sqrt{\frac{I}{MH}}$

For the same magnet, $I$ and $M$ are constant
where $M=$ magnetic moment,
$I=$ moment of inertia of magnet.

$\Rightarrow T\propto \frac{1}{\sqrt{H}}$

first case

$T_1=\frac{60}{40}=\frac{3}{2}=1.5sec.$

$H_1=0.1\times {10}^{-5}T,T_2=2.5sec,H_2=?$

$\Rightarrow \frac{1.5}{2.5}=\sqrt{\frac{H_2}{H_1}}$

$\Rightarrow \frac{15}{25}=\sqrt{\frac{H_2}{{10}^{-6}}}$

$\Rightarrow H_2={\left(\frac{3}{5}\right)}^2\times {10}^{-6}$

$\Rightarrow H_2=\frac{9}{25}\times {10}^{-6}=0.36\times {10}^{-6}T$