Question

A magnet, freely suspended in a vibration magnetometer makes 40 oscillations per minute at a place A and 20 oscillations per minute at B.If B${}_H$ at B is 36$\times$10${}^{-6}$ T, then its value at A is :

• A. $36\times$ 10${}^{-6}$ T
• B. $72\times$ 10${}^{-6}$ T
• C. $144\times$ 10${}^{-6}$ T
• D. $288\times$ 10 ${}^{-6}$ T

Answer 1

The correct option is C $144\times$ 10${}^{-6}$ T

$T=2\pi \sqrt{\frac{I}{MB}}$
So the frequency of oscillation is directly proportional to the square root of magnetic field
$\frac{f_2^2}{f_1^2}=\frac{B_2}{B_1}$
$B_1=\frac{f_1^2}{f_2^2}\times B_2$
$B_1=2^2\times 36\times {10}^{-6}T$
$B_1=144\times {10}^{-}6T$