The magnetic potential due to a magnetic dipole at a point on its axis distant 40 cm from its centre is found to be 2.4 × 10-5 JA -1 m -1. The magnetic moment of the dipole will be

Question

The magnetic potential due to a magnetic dipole at a point on its axis distant 40 cm from its centre is found to be 2.4 × 10-5 JA -1 m -1. The magnetic moment of the dipole will be (A) 28.6 Am 2 (B) 32.2 Am 2 (C) 38.4 Am 2 (D) None of these

Tardigrade Admin · Accepted Answer

Here, r=40 cm =0.4 m θ=0° (an axial line) V=2.4 × 10-5 J / A - m ; M =? As, V =(μ0/4 π) (M cos θ/r2) ⇒ 2.4 × 10-5 =10-7 (M × 1/(0.4)2) or M =38.4 Am 2

Q. The magnetic potential due to a magnetic dipole at a point on its axis distant $40 c m$ from its centre is found to be $2.4 \times 1 0^{- 5} J A^{- 1} m^{- 1}$ . The magnetic moment of the dipole will be

$28.6 A m^{2}$

$32.2 A m^{2}$

$38.4 A m^{2}$

None of these

Here, $r = 40 c m = 0.4 m$
$θ = 0^{\circ}$ (an axial line)
$V = 2.4 \times 1 0^{- 5} J / A - m; M = ?$
As, $V = \frac{μ _{0}}{4 π} \frac{M c o s θ}{r ^{2}}$
$\Rightarrow 2.4 \times 1 0^{- 5} = 1 0^{- 7} \frac{M \times 1}{( 0.4 ) ^{2}}$
or $M = 38.4 A m^{2}$

Q. The magnetic potential due to a magnetic dipole at a point on its axis distant 40cm from its centre is found to be 2.4×10−5JA−1m−1. The magnetic moment of the dipole will be

28.6Am2

32.2Am2

38.4Am2