Question

A magnetic dipole of magnetic moment $0.72\sqrt{2}\mathrm{A}{\mathrm{m}}^2$ is placed horizontally with the north pole pointing towards east. 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 magnetic dipole, M $=0.72\sqrt{2}\mathrm{A}-{\mathrm{m}}^2$
Horizontal component of the earth's magnetic field, B_H = 18 μT
Let d be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet $\left(B\right)$ on the equatorial line of the dipole is given by,
$$\begin{gathered} B=\frac{{\mu }_0}{4\pi }\frac{M}{d^3} \\ \Rightarrow \frac{4\mathrm{\pi }\times {10}^{-7}}{4\mathrm{\pi }}\times \frac{0.72\sqrt{2}}{d^3}=18\times {10}^{-6} \\ \Rightarrow d^3=\frac{0.72\times 1.414\times {10}^{-7}}{18\times {10}^{-6}} \\ \Rightarrow d^3=0.005656 \\ \Rightarrow d=\sqrt[3]{0.005656} \\ \Rightarrow d\approx 0.2\mathrm{m}=20\mathrm{cm} \end{gathered}$$