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 cm from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in 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.)

We are told that the neutral points lie on the East-West line drawn through the mid-point of the bar magnet. Such points are situated on the equatorial (broadside) line of the magnet. For a short bar magnet the magnetic induction at a point on the equatorial line, at a distance $$r$$ from the centre of the magnet, is given by the standard formula

$$B_{\text{equatorial}}=\frac{\mu_0}{4\pi}\,\frac{M}{r^{3}}$$

where $$M$$ is the magnetic moment of the magnet. This field is directed opposite to the direction of the earth’s horizontal field $$B_H$$. At a neutral point the two fields cancel in magnitude, so we must have

$$B_H = \frac{\mu_0}{4\pi}\,\frac{M}{r^{3}}$$

Now we solve for $$M$$. Rearranging the above equality gives

$$M = B_H \, r^{3}\,\left(\frac{4\pi}{\mu_0}\right)$$

We substitute the numerical values. The distance of the neutral point from the magnet’s centre is given as 30 cm, which in SI units is

$$r = 30\ \text{cm} = 0.30\ \text{m}$$

Cubing this distance, step by step,

$$r^{3} = (0.30\ \text{m})^{3} = 0.30 \times 0.30 \times 0.30 = 0.027\ \text{m}^{3}$$

The horizontal component of the earth’s magnetic field is

$$B_H = 3.6 \times 10^{-5}\ \text{T}$$

The constant $$\frac{\mu_0}{4\pi}$$ is given as $$10^{-7}$$ in SI units, hence

$$\frac{4\pi}{\mu_0} = \frac{1}{\mu_0/4\pi} = \frac{1}{10^{-7}} = 10^{7}$$

Substituting everything into the expression for $$M$$, we get

$$M = \left(3.6 \times 10^{-5}\right)\,\left(0.027\right)\,\left(10^{7}\right)$$

We first multiply the two decimal numbers:

$$3.6 \times 0.027 = 0.0972$$

Next we combine the powers of ten:

$$10^{-5} \times 10^{7} = 10^{2} = 100$$

Finally we multiply these results:

$$M = 0.0972 \times 100 = 9.72$$

So, the magnetic moment is

$$M \approx 9.7\ \text{A m}^2$$

Among the given choices, this value corresponds to Option D (9.7 A m2).

Hence, the correct answer is Option D.