A short bar magnet is placed horizontally along the magnetic N-S direction with its axis along the magnetic E-W direction.The resultant horizontal magnetic induction on its equator at a distance of 20 cm from centre is `(B_(H)=4xx10^(-5)T)`

Options

A. `4.5 Am^(2)`
B. `5.4 Am^(2)`
C. `5.0Am^(2)`
D. `4.0 Am^(2)`

Answer

B

Explanation

`B_(H)=B("axis")=(mu_(0))/(4pi)(2M)/(r^(3))`
`4xx10^(-5)=(10^(-7)xx2xxM)/(27xx10^(-3))`
`therefore M=5.4Am^(2)`