A bar magnet of length 6 cm is kept vertically with its north pole on the ground . If the distance of neutral point on the ground is 8 cm from the north pole , what will be the magnetic moment of that magnet ? [H=0.36 CGS units]

Explanation

Let O be the neutral point [fig]
NS=6cm,NO=8cm
`" so, "So=sqrt(6^(2)+8^(2))=10cm`
If the pole -strength of the magnet NS is `q_(m)` magnetic intensity at the point O due to the north pole ,
`H_(1)=(q_(m))/(ON^(2)),` in the direction of OA [in CGS]
Again , due to the south pole magnetic intensity atthe point O ,
`H_(2)=(q_(m))/(OS^(2)),`in the direction of OS ,
`therefore` Component of `H_(2)` in the direction ON,
`H_(2)costheta=(q_(m))/(OS^(2)).(ON)/(OS)`
So , the horizontal magnetic intensity at the point O due to the entire magnet ,
`H_(1)-H_(2)costheta=(q_(m))/(ON^(2))-(q_(m))/(OS^(2)).(ON)/(OS)=q_(m)((1)/(8^(2))-(1)/(10^(2))-(1)/(8))`
= `q_(m)xx0.007625Oe`
Since , O is the neutral point , the magnetic intensity at that point due to the magnet will be equal but opposite to the horizontal component H of earth's magnetic field .
`therefore" " q_(m)xx0.007625=0.36" or, "q_(m)=(0.36)/(0.007625)` emu . cm
`therefore` Magnetic moment of the magnet
`=q_(m).NS=(0.36)/(0.007625)xx6=283.3"emu. cm"^(2)`