Question

A magnetic plane of magnetic moment $1.44A{m}^2$ 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 $18\mu T$

Answer 1

We know for a magnetic dipole with its north pointing the north, the nuetral point in the broadside on position.

Again $\overrightarrow{B}$ in this case $=\frac{{\mu }_{0}M}{4\pi d^3}$

$\therefore \frac{{\mu }_{0}M}{4\pi d^3}={\overrightarrow{B}}_H$ due to earth

$\Rightarrow \frac{{10}^{-7}\times 1.44}{d^3}=18\mu T$

$\Rightarrow \frac{{10}^{-7}\times 1.44}{d^3}=18\times {10}^{-6}$

$\Rightarrow d^3=8\times {10}^{-3}$

$\Rightarrow d=2\times {10}^{-1}m=20cm$

In the plane bisecting the dipole.