A bar magnet `30cm` long is placed in magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of `30cm` from its centre. Calculate the pole strength of the magnet. Given horizontal component of earth's field is `0*34G`.

Explanation

`Here, 2l=30cm or l=15cm=0.15cm`
`" " r=30cm=0.15m,`
`and " " H=0.34G=0.34xx10^(-4)T`
When magnet is placed with its north pole pointing south, then neutral point is obtained on its axial line.
` therefore " " B_("axial")=H or (mu_(0))/(4pi)xx(2Mr)/((r^(2)-l^(2)))=H`
`or " " M=(4pi)/(mu_(0))xx(H(r^(2)-l^(2))^(2))/(2r)`
` " " =(1)/(10^(-7))xx(0.34xx10^(-4)xx(0.30^(2)xx0.15^(2))^(2))/(2xx0.30)`
` " " (0.34xx10^(-4)xx(0.00675)^(2))/(10^(-7)xx2xx0.30)=2.582A-m^(2)`
The pole strength of the magnet,
`" " m=(M)/(2l)=(2.582)/(0.30)=8.606A-m`