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A 6cm long bar magnet possesing magentic...

A `6cm` long bar magnet possesing magentic dipole moment `0*3Am^2` is placed vertically on a horizontal wooden table with north pole of magnet touching the table. A neutral point is found on the table at a distance of `8cm` south of the magnet. Calculate the horizontal component of earth's magnetic field.

Text Solution

Verified by Experts

Here, `2l=6cm`, `l=3cm=3xx10^-2m`
`M=0*3Am^2`, `d=8cm=8xx10^-2m`
`H=?`

In figure.
field due to N pole, `B_N=(mu_0m)/(4pid^2)`, along NP
field due to S pole, `B_S=(mu_0m)/(4pi(d^2+4l^2))`, along PS
Horizontal component of `B_S` is `B_S cos theta`
`=(mu_0m)/(4pi(d^2+4l^2))xx(d)/(sqrt(d^2+4l^2))`
Resultant horizontal field at P
`=B_N-B_Scostheta`
`=(mu_0m)/(4pi)[1/d^2-(d)/((d^2+4l^2)^*(3//2))]`
`=(mu_0M)/(4pi(2l))[1/d^2-(d)/((d^2+4l^2)^(3//2))]`
At neutral point, the resultant horizontal field is balanced by horizontal component of earth's magnetic field H, i.e.,
`H=(mu_0)/(4pi)(M)/((2l))[1/d^2-(d)/((d^2+l^2)^(3//2))]`
`=10^-7xx(0*3)/(6xx10^-2)`
`[(1)/((8xx10^-2)^2)-(8xx10^-2)/([(8xx10^-2)^2+(3xx10^-2)^2]^(3//2))]`
`H=3*81xx10^-5T`
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