Moment of inertia of a megnetic needle is `40 gm-cm^(2)` has time period 3 seconds in earth's horizontal field `=3.6xx10^(-5) weber//m^(2)`. Its magnetic moment will be

Options

A. `0.5 Axxm^(2)`
B. `5 Axxm^(2)`
C. `0.250Axxm^(2)`
D. `5xx10^(2) Axxm^(2)`

Answer

A

Explanation

`T=2pisqrt(I/(MB_(H)))`
`I=40 gm-cm^(2)=400xx10^(-8) kg-m^(2)`
`:. 3=2pisqrt((400xx10^(-8))/(36xx10^(-6)xxM))`
`implies 1/M=9/(4pi^(2))xx36/4 impliesM=0.5 Axxm^(2)`