Question

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is $2$ sec. When another magnet is brought near and parallel to it, then the time period reduces to $1$ sec. The ratio of the horizontal component of earth's field and the field due to magnet will be

• A. $\frac{1}{3}$
• B. $3$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{1}{3}$

Using the below formula gives,
$\frac{B_H}{B}=\frac{1}{3}$
$T=2\pi \sqrt{\frac{I}{M{B}_H}}=2$
$T=2\pi \sqrt{\frac{I}{M(B_H+B)}}=1$
Hence , $\frac{B+B_H}{B_H}=\frac{4}{1}$
$B=3B_H$