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

Correct Answer: 6

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: \(M = 6 \, \text{Am}^2\)