Question

The magnetic field at the centre of a circular current carrying loop of radius 12.3 cm is 6.4 × 10^-6 T. What is the magnetic moment of the loop?

Answer 1

Data: R = 12.3 cm = 12.3 × 10^-2 m,

B = 6.4 × 10^-6 T, µ₀ = 4π x 10^-7 T·m/ A

B = ${\displaystyle \frac{{\mu }_0\text{I}}{\text{2R}}=\frac{{\mu }_0\text{I}}{\text{2R}}\times \frac{2\pi {\text{R}}^2}{2\pi {\text{R}}^2}=\frac{{\mu }_0}{4\pi }\frac{2\text{I}\left(\text{A}\right)}{{\text{R}}^3}}$     (∵ A = πR²)

${\displaystyle =\frac{{\mu }_0}{4\pi }\frac{2\mu }{{\text{R}}^3}}$

where µ = IA is the magnetic dipole moment of the coil.

∴ ${\displaystyle \mu =\frac{{\text{BR}}^3}{2({\mu }_0/4\pi )}=\frac{(6.4\times {10}^{-6}){\left(0.123\right)}^3}{2\times {10}^{-7}}}$

= 5.955 × 10^-2 J/T (or A m²)