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 $2.5s$ 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

Answer: 0.36

$T\propto \frac{1}{\sqrt{B_H}}$
$\Rightarrow \frac{B_{H_2}}{B_{H_1}}={\left(\frac{T_1}{T_2}\right)}^2$
$T_1=\frac{1}{f_1}=\frac{60}{40}=1.5s$
$\frac{B_{H_2}}{B_{H_1}}={\left(\frac{1.5}{2.5}\right)}^2={\left(\frac{3}{5}\right)}^2$
$\Rightarrow B_{H_2}=\frac{9}{25}\times 0.1\times 10^{-5}T$
$=0.36\times 10^{-6}T$