A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole-strength of 200 Am and has a length of 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.

Explanation

Here, `p= 200` Am
`z= 20 cm = 0.2`m
`2l = 5 cm = 0.05 m`
`p_1 = 100` Am
`F=?`
The magnetic moment of bar magnet
`m= 2pl = p(2l)`
`= 200 xx 0.05 = 10 "Am"^2`
The magnetic field at the pole of bar magnet,
`B= (mu_0)/( 4pi ) (2m)/( z^3)`
`B= (2m)/(z^3) xx 10^(-7)`
`= (2 xx 10 xx 10^(-7) )/( (0.2) ^(3) )`
`therefore B= 2.5 xx 10^(-4)` T
Now the force on the magnetic pole
`p_(1) = 100` Am
`F= p_(1) B`
`= 100 xx 2.5 xx 10^(-4)`
`therefore F= 2.5 xx 10^(-2) N`