A bar magnet of length 8 cm is placed on a horizontal plane in the magnetic with its north pole pointing north . If the magnetic moment moment of the the magnet be 90 CGB units and the horizontal component of earth's magnetic field be 0.35 Oe , determine the positions of the neutral points .

Explanation

Let each of two neutral points be at a distance d from the centre of the magnet of the magnet along its perpendicular bisector . If the magnet ic moment of the magnet be `p_(m)` and its magnetic length be 2l , the magnetic intensity at the neutral points due to the magnet (in CGS system) is ,
`H_(m)=(p_(m))/((d^(2)+l^(2))^(3//2))`
At the neutral points , the horizontal component H of earth's magnetic field will be its equal and opposite . ltbtgt `therefore" " H=H_(m)=(p_(m))/(d^(2)+l^(2))^(2//3)" or, " (d^(2)+l^(2))^(3//2)=(p_(m))/(H)`
or , `d^(2)+l^(2)=((p_(m))/(H))^(2//3) "or , " d=sqrt(((p_m)/(H))^(2//3)-l^(2))`
Here , `p_(m)=90"CGS inits , " H=0.35Oe , 2l = 8 "cm i .e, "l = 4 "cm"`
`therefore" " d=sqrt(((90)/(0.35))^(2//3)-4^(2))=4.94"cm"`