Question

A DMM is placed with its arms in the N−S direction. The distance at which a short bar magnet having $\dfrac{M}{{{B_H}}} = 80\dfrac{{A{m^2}}}{T}$ 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

According to the question we have to find the value of distance $d$ at which a short bar magnet should be placed. Using the formula which gives the relation between the magnetic dipole moment ( $M$ ) of a bar magnet and horizontal intensity ( ${B_H}$ ) of earth’s magnetic field using a deflection magnetometer. We will answer this question.
$\dfrac{M}{{{B_H}}} = \dfrac{{4\pi }}{{{\mu _0}}}\dfrac{{{{\left( {{d^2} - {l^2}} \right)}^{\dfrac{3}{2}}}}}{{2d}}\tan \theta$
Where $M$ is the magnetic dipole moment, ${B_H}$ is the intensity of the magnet, $d$ is the distance at which the magnet is placed, $l$ is the length of the magnet, $\dfrac{{4\pi }}{{{\mu _0}}}$ is a constant, $\theta$ is the angle.

Complete answer:
The horizontal component of the earth's magnetic field, denoted by ${B_H}$ is the component of the earth's magnetic field along a horizontal plane whose normal vector runs through the earth's centre.
A magnetic dipole's magnetic dipole moment is the attribute of the dipole that causes it to align parallel to an external magnetic field.
Using the formula
$\dfrac{M}{{{B_H}}} = \dfrac{{4\pi }}{{{\mu _0}}}\dfrac{{{{\left( {{d^2} - {l^2}} \right)}^{\dfrac{3}{2}}}}}{{2d}}\tan \theta$
For this question the angle is not considered as it is given that the needle can stay in any position so
$\Rightarrow \dfrac{M}{{{B_H}}} = \dfrac{{4\pi }}{{{\mu _0}}}\dfrac{{{{\left( {{d^2} - {l^2}} \right)}^{\dfrac{3}{2}}}}}{{2d}}$
Also for short bar magnet $d > > l$
$\Rightarrow \dfrac{M}{{{B_H}}} = \dfrac{{4\pi }}{{{\mu _0}}}\dfrac{{{d^3}}}{2}$
$\Rightarrow {d^3} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 80}}{{4\pi }}$
$d = 2 \times {10^{ - 2}} = 2cm$

The value is positive so it is placed in the Gaussian south.
Hence option D) is the correct answer.

Note:
The deflection magnetometer is made up of a big compass box with a small magnetic needle pivoting at the centre of a circular scale so that it can rotate freely in a horizontal plane. The magnetic needle is tightly placed perpendicular to a huge aluminum pointer.