A small bar magnet has a magnetic moment `5Am^2`. The neutral point is obtained on axial line when it is placed in magnetic meridian with its N pole pointing S of earth and neutral point is obtained on equatorial line, when it is placed with its N pole pointing towards north of earth. If horizontal component of earth's field is `0*38G`, find the position of neutral points in the two cases.

Explanation

Here, `M=5Am^2`,
`H=0*38G=0*38xx10^-4T`.
If `d_1` is distance of neutral point when N pole of magnet is pointing S of earth, then
`B_1=(mu_0)/(4pi)(2M)/(d_1^3)=H`
`d_1^3=(mu_0)/(4pi)(2M)/(H)=10^-7xx(2xx5)/(0*38xx10^-4)`
`=26*3xx10^-3m`
`d_1=(26*3xx10^-3)^(1//3)m=2*97xx10^-1m`
`=29*7cm`
In the second case, `B_2=(mu_0)/(4pi)(M)/(d_2^3)=H`
`d_2^3=(mu_0)/(4pi)M/H=(10^-7xx5)/(0*38xx10^-4)=13*15xx10^-3`
`d_2=(13*15xx10^-3)^(1//3)m=2*36xx10^-1m`
`=23*6cm`