1. Magnetic scalar potential on the axis of a dipole

For a magnetic dipole of moment $M$, the magnetic potential at a point at distance $r$ making angle $\theta$ with the dipole axis is

$V = \frac{\mu_0}{4\pi}\frac{M\cos\theta}{r^2}$

On the axis of the dipole, $\theta = 0$ so $\cos\theta = 1$. Hence,

$V_{\text{axis}} = \frac{\mu_0}{4\pi}\frac{M}{r^2}$

2. Substitute the given values

Given:

• $V = 1.5 \times 10^{-5}\ \text{T m}$
• $r = 20\ \text{cm} = 0.2\ \text{m}$
• $\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$

Using

$1.5 \times 10^{-5} = 10^{-7}\cdot \frac{M}{(0.2)^2}$

3. Solve for magnetic moment $M$

First,

$(0.2)^2 = 0.04$

So,

$1.5 \times 10^{-5} = 10^{-7}\cdot \frac{M}{0.04}$

$M = \frac{1.5 \times 10^{-5} \times 0.04}{10^{-7}}$

$M = 1.5 \times 0.04 \times 10^2$

$M = 0.06 \times 100 = 6$

4. Final answer

$\boxed{6\ \text{A m}^2}$

5. Comparison with stored answer

Stored correct answer = $6$

Our derived answer also is $6$, so they agree.