Physics - Magnetism Question with Solution | TestHub

The magnetic field at the centre of a circular coil is given by:

$B=\frac{{\mu }_{0}NI}{2R}$

Substituting the given values ($B=3.14\times 10^{-3}$ T, $N=100$, $R=0.05$ m, ${\mu }_0=4\pi \times 10^{-7}$ T m/A):

$3.14\times 10^{-3}=\frac{4\pi \times 10^{-7}\times 100\times I}{2\times 0.05}$

Taking $\pi =3.14$:

$3.14\times 10^{-3}=\frac{4\times 3.14\times 10^{-5}\times I}{0.1}$

$10^{-3}=4\times 10^{-4}\times I$

$I=\frac{10^{-3}}{4\times 10^{-4}}=2.5\text{ A}$

The magnetic moment of the coil is given by:

$M=NIA=NI(\pi R^2)$

$M=100\times 2.5\times 3.14\times (0.05{)}^2$

$M=250\times 3.14\times 0.0025$

$M=1.9625{\text{ A m}}^2\approx 2{\text{ A m}}^2$

Answer: $2.5$ A, $2$ A m${}^2$