Question

A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is seconds. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is

• A. $2^{\frac{1}{4}}$
• B. $2^{\frac{1}{2}}$
• C. $2$
• D. $2^{\frac{3}{4}}$

Answer 1

The correct option is C $2$

Initially magnetic moment of system
$M_1=\sqrt{M^2+M^2}=2M$ and moment of inertia
$I_1=I+I=2I$
Finally when one of the magnet is removed then
$M_2=Mand{I}_2=I$
So $T=2\pi \sqrt{\frac{I}{M{B}_H}}$
$\frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}\times \frac{M_2}{M_1}}=\sqrt{\frac{2I}{I}\times \frac{M}{\sqrt{2}M}}\Rightarrow T_2=\frac{2^{\frac{5}{4}}}{2^{\frac{1}{4}}}=2sec$