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

The position of the neutral point when the horizontal component of the earth's magnetic field is 18 μT is 20 cm

Explanation:

Given data in the question

The Magnetic Dipolar Moment, $M=1.44 A-m^{2}$

Horizontal part of a magnetic field on Earth, $B_{H}=18 \mu T$

We know that the neutral point always lies in the broadside-on position for a magnetic dipole with pole pointing toward the north.

If d be the center of the magnet parallel to the neutral point.

Because of the dipole, the magnetic area at the broadside on place (B) is given by

$\vec{B}=\frac{\mu_{0} M}{4 \pi d^{3}}$

At that point, this strength of the magnetic area should also be equal to horizontal portion of the Magnetic layer of Earth, that is, due to Earth.

$\begin{aligned}&\vec{B}=\frac{\mu_{0} M}{4 \pi d^{3}}\\&B=\frac{10^{-7} \times 1.44}{d^{3}}\\&18 \times 10^{-6}=\frac{10^{-7} \times 1.44}{d^{3}}\\&d^{3}=8 \times 10^{-3}\\&d=2 \times 10^{-1} \mathrm{m}=0.2 \mathrm{m}=20 \mathrm{cm}\end{aligned}$