1.

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)

Answer»

`4.5 Am^(2)`
`5.4 Am^(2)`
`5.0Am^(2)`
`4.0 Am^(2)`

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


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