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

Time period in vibration magnetometer
`T=2pisqrt(I/(MxxB_(H))) impliesT prop1/sqrt(B_(H))`
`implies T_(1)/T_(2)=sqrt((B_(H))_(2)/((B_(H))_(1)))implies 2/T_(2)=sqrt(18/24)`
`T=2.3 s ~~2 s`