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 6 microtesla. When a horizontal field of 18 microtesla is produced in same direction of field by placing a current carrying wire, the new time period of magnet will be :-

Question

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 6 microtesla. When a horizontal field of 18 microtesla is produced in same direction of field by placing a current carrying wire, the new time period of magnet will be :- (A) 4s (B) 1s (C) 2s (D) 3s

Tardigrade Admin · Accepted Answer

T =2 π √( I / MB ) ⇒ T propto (1/√ B ) ( T 1/ T 2)=√( B 2/ B 1)=√(6+18/6)=2 ⇒ T 2=1 sec

4s

1s

2s

3s

T=2πMBI​​ ⇒T∝B​1​ T2​T1​​=B1​B2​​​=66+18​​=2 ⇒T2​=1sec

$T = 2 π \frac{I}{MB}$
$\Rightarrow T \propto \frac{1}{B}$
$\frac{T _{1}}{T _{2}} = \frac{B _{2}}{B _{1}} = \frac{6 + 18}{6} = 2$
$\Rightarrow T_{2} = 1 sec$