Question

A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having M/BH = 40 A m2 T−1 be placed so that the needle can stay in any position?

Answer 1

The magnet should therefore be positioned in such a way that its north pole points southwards and is 2 cm away from the needle.

Explanation:

Given data in the question  

Ratio of M and  $B_H$

$\frac{M}{B_{H}}=40 \mathrm{Am}^{2} / \mathrm{T}$

Since the magnet is small, it is possible to ignore l.

Then, use the formula for $\frac{M}{B_{H}}$ from the principle of the magnetometer, and replacing both values, we get

$\frac{M}{B_{H}}=\frac{4 \pi}{\mu_{0}} \frac{d^{3}}{2}=40$

$d^{3}=40 \times 10^{-7} \times 2$

$d^{3}=8 \times 10^{-6}$

  $d=2 \times 10^{-2} \mathrm{m}$

     $=2 \mathrm{cm}$

The magnet should therefore be positioned in such a way that its north pole points southwards and is 2 cm away from the needle.