A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of \(3.14 \times 10^{-3}\) T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively :
(Take \(\mu_0=4 \pi \times 10^{-7} T m / A\) )

(D) \(2.5 A, 2 A m ^2\)
The magnetic field at the centre of a circular coil is given by:
\(B=\frac{\mu_0 N I}{2 R}\)
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 \(\pi=3.14\) :
\(3.14 \times 10^{-3}=\frac{4 \times 3.14 \times 10^{-5} \times I}{0.1}\)
\(10^{-3}=4 \times 10^{-4} \times I\)
\(I=\frac{10^{-3}}{4 \times 10^{-4}}=2.5 A\)
The magnetic moment of the coil is given by:
\(M=N I A=N I\left(\pi R^2\right)\)
\(M=100 \times 2.5 \times 3.14 \times(0.05)^2\)
\(M=250 \times 3.14 \times 0.0025\)
\(M=1.9625 A m^2 \approx 2 A m^2\)