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A bar magnet of length 8 cm and having a...

A bar magnet of length `8 cm` and having a pole strengh of `1.0 A m` is placed vertically on a horizontal table with its south pole on the table. A neutral point is found on the table at a distance of 6.0 cm north of the magnet. Calculate the earth's horizontal magnetic field.

A

`22xx10^(-6)T`

B

`44x10^(-6)T`

C

`66xx10^(-6)T`

D

`88xx10^(-6)T`

Text Solution

Verified by Experts

The correct Answer is:
A
The magnetic field at `P` due to south pole
`B_(s)=(mu_(0))/(4pi)(m)/(r^(2))` and de to north pole
`B_(N)=(mu_(0))/(4pi)(m)/((l^(2)+r^(2)))` along `NP`
Its component along north is `B_(N) cos theta`. On substituting the value of `theta`, it
`(mu_(0))/(4pi)(m)/((l^(2)+r^(2))^(3//2))xx(r )/(sqrt(l^(2)+r^(2))`
`(mu_(0))/(4pi)(m)/((l^(2)+r^(2))^(3//2))`
For neutral point at `P`, we have
`B_(H)=B_(S)-B_(N) cos theta` `=(mu_(0))/(4mu)[(m)/(r^(2))(mr)/((l^(2)+r^(2))^((3)/(2)))]`
`=10^(-7)x1[(1)/(0.06^(2))-(0.06)/((0.08^(2)+0.06^(2))^((3)/(2)))]`
`=22xx10^(-6)T`
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