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 × 10⁻⁵ T m. The magnetic moment of the dipole is _________ A m². (Given : μₒ/(4 π)=10⁻⁷ Tm A⁻¹ )

Solution

Physics sir · 4 steps

On the axis, magnetic potential falls as 1/r², so multiply by r² and divide by μ₀/4π.

1. Axial potential. For a short magnetic dipole of moment M, the potential at distance r on its axis has the same form as the electric dipole case, with μ₀/4π in place of 1/4πε₀.

   V = μ₀/4π M/r² ⇒ M = (V r²)/(μ₀/4π)

2. Convert to metres. r = 20 cm = 0.2 m, so r² = 0.04 m².
3. Put in numbers. M = (1.5 × 10⁻⁵ × 0.04)/10⁻⁷ = (6 × 10⁻⁷)/10⁻⁷ = 6 A m².
4. So. The magnetic moment of the dipole is 6 A m².

The common slip: 60000. Leaving r in centimetres, so that r² = 400, gives 60000, not 6. Convert 20 cm to 0.2 m first.

The idea to keep. Dipole potential goes as 1/r² and dipole field goes as 1/r³. Pick the right power, and always put r in metres before you square it.

Answer from the NTA final answer key. Solution by PrepWiser.