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

Answer

d

Explanation

(d) Time period of a vibration magnetometer,
`Tprop (1)/(sqrtB) rArr (T_(1))/(T_(2))=sqrt((B_(2))/(B_(1)))`
`rArrT_(2)=T_(1) sqrt((B_(1))/(B_(2)))=2sqrt((24xx10^(-6))/(6xx10^(-6)))=4s`