Question

The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?

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,
$$\begin{gathered} \frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\frac{{\left(d^2-l^2\right)}^2}{2d}\tan \theta \\ \mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet}, \\ \frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\times \frac{d^4}{2d}\tan \theta \\ \Rightarrow \frac{M}{B_H}=\frac{4\mathrm{\pi }}{4\mathrm{\pi }\times {10}^{-7}}\times \frac{{\left(0.1\right)}^3}{2}\times \tan 37° \\ \Rightarrow \frac{M}{B_H}=0.5\times 0.75\times 1\times {10}^4 \\ \Rightarrow \frac{M}{B_H}=3.75\times {10}^3\mathrm{A}-{\mathrm{m}}^2/\mathrm{T} \end{gathered}$$
= 3.75 × 10³ A-m²/T
Deflection in the magnetometer θ = 37°
From the magnetometer theory in Tan-B position, we have
$\frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}{\left(d^2+l^2\right)}^{3/2}\tan \mathit{}\theta$
Since for the short magnet l << d, we can neglect l w.r.t. d.
Now,
$$\begin{gathered} \frac{M}{B_H}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}{d}^3\tan \theta \\ \Rightarrow 3.75\times {10}^3=\frac{1}{{10}^{-7}}\times d^{\mathit{3}}\times 0.75 \\ \Rightarrow d^3=\frac{3.75\times {10}^3\times {10}^{-7}}{0.75}=5\times {10}^{-4} \\ \Rightarrow d=\sqrt[3]{5\times {10}^{-4}} \\ \Rightarrow d=0.079\mathrm{m}=7.9\mathrm{cm} \end{gathered}$$
Magnet will be at 7.9 cm from the centre.