Q: 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 2^{(5/4)} sec. 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^{(1/4)}

(b) 2^{(1/2)}

(c) 2

(d) 2^{(3/4)}

Ans: (c)

Sol: When two identical bar magnets are held perpendicular to each other.

$ \displaystyle M_1 = \sqrt{M^2 + M^2}$

$\displaystyle M_1 = \sqrt{2} M $ ; I_{1} = I

T_{1} = 2^{(5/4)} sec ; T_{2} = ?

M_{2} = M (as one magnet is removed)

I_{2} = I/2

$\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{I_2}{I_1}\times \frac{M_1}{M_2}} $

$\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{1}{2}\times \frac{\sqrt{2}M}{M}} $

$ \displaystyle \frac{T_2}{T_1} = \frac{1}{2^{1/4}} $

$\displaystyle T_2 = \frac{T_1}{2^{1/4}} $

$ \displaystyle T_2 = \frac{2^{5/4}}{2^{1/4}} $

T_{2} = 2 sec