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.

Options

A. `1.25`
B. `2.5`
C. `5.0`
D. `1.8`

Answer

B

Explanation

`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`