Question

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 m² T⁻¹ be placed so that the needle can stay in any position?

Answer 1

Given:
Ratio, $\frac{M}{B_H}=40{\mathrm{Am}}^2/\mathrm{T}$
Since the magnet is short, l can be neglected.
So, using the formula for $\frac{\mathit{M}}{{\mathit{B}}_{\mathit{H}}}$ from the magnetometer theory and substituting all values, we get
$$\begin{gathered} \mathit{}\frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\frac{d^3}{2}=40 \\ \Rightarrow d^3=40\times {10}^{-7}\times 2 \\ \Rightarrow d^3=8\times {10}^{-6} \\ \Rightarrow d=2\times {10}^{-2}\mathrm{m}=2\mathrm{cm} \end{gathered}$$

Thus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.