Question

A magnet makes $40$ oscillations per minute at a place having magnetic field of $0.1 \times 10^{-5} T$. At another place, it take $2.5\, s$ to complete one vibration. If the value of the earth's horizontal field at that place is $y \times 10^{-6} T$. then find $y$.

Answer 1

$T \propto \frac{1}{\sqrt{B_{H}}} $
$\Rightarrow \frac{B_{H_{2}}}{B_{H_{1}}}=\left(\frac{T_{1}}{T_{2}}\right)^{2}$
$T_{1}=\frac{1}{f_{1}}=\frac{60}{40}=1.5 \,s $
$\frac{B_{H_{2}}}{B_{H_{1}}}=\left(\frac{1.5}{2.5}\right)^{2}=\left(\frac{3}{5}\right)^{2}$
$\Rightarrow B_{H_{2}}=\frac{9}{25} \times 0.1 \times 10^{-5} \,T $
$=0.36 \times 10^{-6} \,T$