A very short bar magnet has magnetic moment `1*4175JT^-1`. It is placed (i) with its north pole pointing towards geographic north (ii) with its north pole pointing towards geographic south. If horizontal component of earth's field at the place is `0*42gauss`, calculate the distance of the neutral points from the magnet.

Answer

`15cm`; `18*9cm`

Explanation

(i) `H=(mu_0)/(4pi)M/d^3`
or `d=[(mu_0)/(4pi)M/H]^(1//3)=[(10^-7xx1*4175)/(0*42xx10^-4)]^(1//3)`
`=0*15m=15cm`
(ii) `H=(mu_0)/(4pi)(2M)/(d^3)`
or `d=[(mu_0)/(4pi)(2M)/(H)]^(1//3)=[(10^-7xx2xx1*4175)/(0*42xx10^-4)]^(1//3)`
`=0*189m=18*9cm`