First slide

Magnetism

Question

The magnetic scalar potential due to a magnetic field at a point on its axis situated at a distance of 20 cm from its center is found to be $1.2 \times 10^{- 5} V$ . Then, the magnetic moment of the dipole is

Moderate

A

${4.8 Am}^{2}$
B

$9.6 {Am}^{2}$
C

$2.4 {Am}^{2}$
D

$1.2 {Am}^{2}$

Solution

The magnetic potential due to dipole is $V = \frac{μ_{0}}{4 π} . \frac{M \cos θ}{r^{2}}$

${Given : V=1.2×10}^{-5} Tm$ ; $θ {=0}^{0}; r = 0.2 m$
$∴ 1.2 \times 10^{- 5} = (10^{- 7}) \frac{M}{{(0.2)}^{2}}$
${⇒M=4.8 Am}^{2}$ .

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