A planet has magnetic dipole moment of $27\times {10}^{22} \mathrm{A} {\mathrm{m}}^2$. If the radius of the planet is $300 \mathrm{km}$, what would be the magnetic field at its equator?
$\left(\text{Take}, \frac{\mu }{4\pi }={10}^{-7}\right)$

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of $30 \mathrm{cm}$ from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in ${\mathrm{Am}}^2$ is close to:

(Given $\frac{{\mu }_0}{4\pi }={10}^{-7}$ in SI units and $B_H=$ Horizontal component of earth's magnetic field $=3.6\times {10}^{-5}$ Tesla.)

A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$-axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2D$ on positive $Y$-axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $\left|\overrightarrow{B}\right|=\alpha \left[\frac{{\mu }_0}{4\pi }\frac{M}{D^3}\right]$. The value of $\alpha$ must be (Assume $D>>$ where $l$ is the length of magnets)