Question

A DMM is placed with its arms in $N-S$ direction.The distance at which a short bar magnet having $\frac{M}{B_H}=80A{m}^2/T$ should be placed, so that the needle can stay in any position is (nearly)

• A. $2.5cm$ from the needle, $N-$pole pointing GS
• B. $2cm$ from the needle, $N-$ pole pointing GN
• C. $4cm$ from the needle, $N-$ pole pointing GN
• D. $2cm$ from the needle, $N-$ pole pointing GS

Answer 1

The correct option is D $2cm$ from the needle, $N-$ pole pointing GS

Here, DMM is placed in $\tan B$ position, we have
$\frac{{\mu }_{0}M}{4\pi d^3}=B_H\tan \theta$
where, variables have their usual meanings.
$d^3\tan \theta =\frac{{\mu }_{0}M}{4\pi B_H}$
$d^3\tan \theta =\frac{4\pi \times {10}^{-7}\times 80}{4\pi }$
$d^3\tan \theta =8\times {10}^{-6}$
$d\tan \theta =2\times {10}^{-2}=2cm$
and the needle is in position with $N-pole$ pointing Gaussian South.