A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Answer

D

Explanation

The time period T of oscillation of magnetic is given by
`T = 2pi sqrt(I)/(MB)`
where,
I= moment of inertia of the magnet about the aixs of rotation
M = Magnetic moment of the magnet
B = Uniform magnetic field
As the I, B remains the same
`therefore T prop (1)/(sqrt(B)) or (T_2)/(T_1) = sqrt(B_1/B_2)`
According to given problem,
`B_1 = 24 mu T`
`B_2 = 24 mu T - 18 mu T = 6 mu t `
`T_1 = 2 S `
`therefore T_1 = (2 s) sqrt((24 mu T))/(6 mu T) - 4 s`