1.

The force acting on a north pole of magnet of pole strength 3200 Am and 10 cm away from the south pole of a point bar magnet of pole strength 40 Am is .......... N.

Answer»

`-1.28`
`1.28`
`1.28 xx 10^(-7)`
`1.28 xx 10^(-7)`

Solution :`F= (mu_0)/( 4PI ) . (p_1 p_2)/( r^2)`
`= 10^(-7) xx (3200 xx 40)/((0.1)^(2))`
`=12800000 xx 10^(-7) `
`therefore F=1.28 N`


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