If at a distance of 40 cm at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is 2.4× 10- 5 J A m- 1 , then the magnetic moment of the dipole is

Question

If at a distance of 40 cm at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is 2.4× 10- 5 J A m- 1 , then the magnetic moment of the dipole is (A) 28.6Am2 (B) 32.2Am2 (C) 38.4Am2 (D) None of these

Tardigrade Admin · Accepted Answer

Here, r=40 cm=0.4 m θ =0° (an axial line) M=? As V=(μ 0/4 π )(M c o s θ /r2) ⇒ 2.4 × 10-5=10-7 × (M × 1/(0.4)2) ⇒ M=38.4Am2

Q. If at a distance of $40 c m$ at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is $2.4 \times 1 0^{- 5} J A m^{- 1}$ , then the magnetic moment of the dipole is

$28.6 A m^{2}$

$32.2 A m^{2}$

$38.4 A m^{2}$

None of these

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

Q. If at a distance of 40cm at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is 2.4×10−5JAm−1 , then the magnetic moment of the dipole is

28.6Am2

32.2Am2

38.4Am2