Question

A magnetic dipole of magnetic moment 1.44 A m² is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 μT.

Answer 1

Given:
Magnetic moment of the magnetic dipole, M = 1.44 A-m²
​Horizontal component of Earth's magnetic field, B_H = 18 μT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B) is given by
$\overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu }}_{0}M}{4\mathrm{\pi }{d}^3}$
This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, B_H due to Earth.
Thus,
$$\begin{gathered} \overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu }}_{0}M}{4\mathrm{\pi }{d}^3} \\ \Rightarrow B=\frac{{10}^{-7}\times 1.44}{d^3} \\ \Rightarrow 18\times {10}^{-6}=\frac{{10}^{-7}\times 1.44}{d^3} \\ \Rightarrow d^{\mathit{3}}=8\times {10}^{-3} \\ \Rightarrow d=2\times {10}^{-1}=20\mathrm{cm} \end{gathered}$$