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} $)

Question

The magnetic susceptibility $\chi$ for superconductors is ______.

Answer 1

1. Magnetic Potential on Dipole Axis Formula:

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

Given: 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: 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:

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

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

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

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: \( \frac{\mu_0}{4\pi} = 10^{-7} \, \text{TmA}^{-1} \))