Question

A vibration magnetometer consists of two identical bar magnets placed one over the other, such that they are mutually perpendicular and bisect each other. The time period of oscillations of combination in a horizontal magnetic field is $4s$. If one of the magnets is removed, then the period of oscillations of the other in the same field is

• A. $2^{1/4}sec$
• B. $2^{7/4}sec$
• C. $2^{5/4}sec$
• D. $2^{3/4}sec$

Answer 1

Answer: (B)

$2^{7/4}sec$

Given that the time period of oscillation of the combination in horizontal magnetic field is 4 s .

Time period of oscillation in vibration magnetometer is proportional to moment of inertia and inversely proportional to effective magnetic moment of the magnet.

$T\propto \sqrt{\frac{I}{L}}$

Let Initial Moment of Inertia and Magnetic Moment of the combination are I and M respectively .

If one of the magnets is removed , the resultant magnetic moment will be $\frac{M}{\sqrt{2}}$

Similarly , Moment of Inertia after removing one magnet is $\frac{I}{2}$ ( as moment of inertia passing through midpoint of magnets along perpendicular plane, is directly added for two magnets )

Hence ,

$\frac{T_1}{T_2}=\sqrt{\frac{I_1{M}_2}{I_2{M}_1}}$

$=\sqrt{\frac{I\frac{M}{\sqrt{2}}}{M\frac{I}{2}}}=2^{1/4}$

Hence , $T_2=\frac{T_1}{2^{1/4}}$ sec

$T_2=\frac{4}{2^1/4}$ sec

$T_2=2^{7/4}$ sec