1.

A magnet of magnetic moment 0.1 Am^(2) is placed in a uniform magnetic field 0.36 xx 10^(-4) T. The force acting on its each pole is 1.44 xx 10^(-4) N. The distance between two poles would be ...... cm.

Answer»

`1.25`
`2.5`
`5.0`
`1.8`

SOLUTION :`F= PB`
`therefore p =(F)/( B) = (1.44 xx 10^(-4))/( 0.36 xx 10^(-4) ) ""…(1)`
`therefore p = 4 Am`
Now `m= 2pl`
`therefore 2L = (m)/( p) ""…(2)`
`= (0.1)/(4)`
`=0.025` m
`=2.5 cm`


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