A short bar magnet of magnetic moment `3.2 Am^(2)`is placed on a horizontal table with its axis along the magnetic E-W direction.The resultant horizontal magnetic induction on its equator at a distance of 20 cm from its centre is`(B_(H)=4 xx10^(-5) T)`

Options

A. `5.6xx10^(-5) T`
B. `6.5xx10^(-5)T`
C. `5.6xx10^(5)T`
D. `6.5xx10^(5)T`

Answer

A

Explanation

`B_(eq)=(mu_(0))/( 4pi)(M)/(r^(3))=(10^(-7)xx3.2)/(8xx10^(-3))=4xx10^(-5)J`
The resultant horizontal magnetic induction on its equator is given by
`B=sqrt(B_(H)^(2)+B_(eq)^(2))`
`=sqrt((4xx10^(-5))^(2)+(4xx10^(-5))^(2))=5.656xx10^(-5)T`