A deflection megnetometer is placed with its arms in north-south direction. How and where should a short magnet having `M/B_H = 40 A m^2 T(-1)` be placed so that the needle can stay in any position?

Explanation

Given `(M)/(B_(H)) = 40 (A- m^(2))/(T)`
Since the magnet is short l can be neglected
`So (M)/(B_(H)) = (4 pi)/(mu_(0)) (d^(2))/(2) = 40`
rArr d^(2) = (40 xx 4 pi xx 10^(-7) xx 2)/(0.75)`
` = 8 xx 10^(-8)`
`d = 2 xx 10^(-2) m = 2 cm`
With the nprth pole pointing toward smooth