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 $1.5 \times 10^{-5} \mathrm{Tm}$. The magnetic moment of the dipole is $\_\_\_\_$ $\mathrm{Am}^{2}$. (Given : $\frac{\mu_{0}}{4 \pi}=10^{-7} \mathrm{TmA}^{-1}$)

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Magnetic Effects of Current

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 1 0^{- 5} Tm

. The magnetic moment of the dipole is

____

Am^{2}

.
(Given :

\frac{μ _{0}}{4 π} = 1 0^{- 7} TmA^{- 1}

)

Answer: 6

Solution

Magnetic potential on the axis of a dipole:

V = \frac{μ _{0}}{4 π} \frac{M}{r ^{2}}

1.5 \times 1 0^{- 5} = (1 0^{- 7}) \frac{M}{( 0.2 ) ^{2}} ⟹ M = \frac{1.5 \times 1 0 ^{- 5} \times 0.04}{1 0 ^{- 7}} = 6 Am^{2}

Key confirmed: 6.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer

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