A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If $$B_H = 0.4$$ G, the magnetic moment of the magnet is (1 G = $$10^{-4}$$ T):

A bar magnet of length $$2l = 14$$ cm (so $$l = 7$$ cm $$= 0.07$$ m) is placed in the magnetic meridian with its north pole pointing towards geographic north. A neutral point is found at a distance $$r = 18$$ cm $$= 0.18$$ m from the center of the magnet.

When the north pole of the magnet points toward geographic north, the magnet's axial field reinforces Earth's horizontal field on both sides along the axis. On the equatorial line (perpendicular bisector), the magnet's field points opposite to the magnet's magnetic moment direction (i.e., southward), opposing the Earth's field. So the neutral points lie on the equatorial line.

At the neutral point on the equatorial line, the magnetic field due to the magnet equals $$B_H$$: $$\frac{\mu_0}{4\pi} \cdot \frac{M}{(r^2 + l^2)^{3/2}} = B_H$$.

Solving for $$M$$: $$M = \frac{B_H \cdot (r^2 + l^2)^{3/2}}{\mu_0 / (4\pi)}$$.

Computing $$r^2 + l^2 = (0.18)^2 + (0.07)^2 = 0.0324 + 0.0049 = 0.0373$$ m$$^2$$.

$$(r^2 + l^2)^{3/2} = (0.0373)^{3/2} = 0.0373 \times \sqrt{0.0373} = 0.0373 \times 0.19314 = 7.204 \times 10^{-3}$$ m$$^3$$.

With $$B_H = 0.4$$ G $$= 4 \times 10^{-5}$$ T and $$\frac{\mu_0}{4\pi} = 10^{-7}$$ T m/A:

$$M = \frac{4 \times 10^{-5} \times 7.204 \times 10^{-3}}{10^{-7}} = \frac{2.8816 \times 10^{-7}}{10^{-7}} = 2.880$$ J T$$^{-1}$$.