Q.The magnetic field at the centre of a circular current-carrying loop of radius 12.3 cm is $6.4\times10^{-6}$ T. What will be the magnetic moment of the loop?

[!TLDR] From $B=\mu_0I/(2R)$ and $m=I\pi R^2$, combining gives $m=\dfrac{2\pi BR^3}{\mu_0}\approx5.954\times10^{-2}$ A.m$ …

The field at the centre of a circular loop of radius R is $B=\dfrac{\mu_0 I}{2R}$ (Section 10.12), so the current is $I=\dfrac{2BR}{\mu_0}$. The magnetic moment is $m=I\pi R^2=\dfrac{2BR}{\mu_0}\cdot\pi R^2=\dfrac{2\pi BR^3}{\mu_0}$. Substituting $B=6.4\times10^{-6}$ T, $R=12.3\ \text{cm}=0.123$ m, $\mu_0=4\pi\times10^{-7}$ T.m/A: $m=\dfrac{2\pi(6.4\times10^{-6})(0.123)^3 …

Eliminate the current between the centre-of-loop field formula $B=\mu_0I/(2R)$ and the magn …

Using the AXIAL field formula (Section 10.13) instead of the simpler centre-of-loop formula for this at-the-centre case …