Question

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is $1.5 \times 10^{-5} \, \text{Tm}$. The magnetic moment of the dipole is _______ $\text{Am}^2$.
(Given: $\frac{\mu_0}{4\pi} = 10^{-7} \, \text{TmA}^{-1}$)

Answer 1

1. Formula for Magnetic Potential on the Axis of a Dipole:

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

- Where: - $V = 1.5 \times 10^{-5} \, \text{Tm}$ - $\frac{\mu_0}{4 \pi} = 10^{-7} \, \text{Tm/A}$ - $r = 20 \, \text{cm} = 0.2 \, \text{m}$

Step 2: Rearrange to Solve for M:

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

Step 3: Substitute Values:

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

$M = \frac{1.5 \times 10^{-5} \times 4 \times 10^{-2}}{10^{-7}}$

Step 4: Simplify Calculation:

$M = \frac{1.5 \times 4 \times 10^{-7}}{10^{-7}}$

$M = 6 \, \text{Am}^2$

So, the correct answer is: $M = 6 \, \text{Am}^2$