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

Answer

A

Explanation

`T=2pisqrt(I/(MB))`
`T_(2)/T_(1)=sqrt(B_(1)/B_(2))=sqrt(24/6)`
`T_(2)=2T_(1)=4s`