Question

A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.

Answer 1

Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and B_H be Earth's horizontal magnetic field.
According to the magnetometer theory,
$$\begin{gathered} \frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\frac{{\left(d^2-l^2\right)}^2}{2d}\tan \theta \\ \mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet}, \\ \frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\times \frac{d^4}{2d}\tan \theta \\ \Rightarrow \frac{M}{B_H}=\frac{4\mathrm{\pi }}{4\mathrm{\pi }\times {10}^{-7}}\times \frac{{\left(0.1\right)}^3}{2}\times \tan 37° \\ \Rightarrow \frac{M}{B_H}=0.5\times 0.75\times 1\times {10}^4 \\ \Rightarrow \frac{M}{B_H}=3.75\times {10}^3\mathrm{A}-{\mathrm{m}}^2/\mathrm{T} \end{gathered}$$