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A deflection magnetometer is placed with...

A deflection magnetometer 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?

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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
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