2026 Moving Charges and MagnetismMagnetic Field due to a Circular Coil Medium Question 1 of 15 in Magnetic Field due to a Circular Coil

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 $\mu_0 = 4\pi \times 10^{-7}$ Tm/A)

A 2.5 A, 2 $Am^2$
B 2.5 A, 20 $Am^2$
C 2 A, 4 $Am^2$
D 2 A, 10 $Am^2$

The answer is A.

Solution

$B = \frac{\mu_0 N I}{2R} \Rightarrow I = \frac{2BR}{\mu_0 N}$ $I = \frac{2 \times 3.14 \times 10^{-3} \times 5 \times 10^{-2}}{4\pi \times 10^{-7} \times 100} = 2.5$ A Magnetic moment $M = NIA = 100 \times 2.5 \times \pi r^2$ $= 100 \times 2.5 \times 3.14 \times 25 \times 10^{-4} = 1.96 \approx 2\ A\,m^2$ So I = 2.5 A and M = 2 $A\,m^2$.

Magnetic Field due to a Circular Coil in past papers

15 questions from this chapter have appeared across 12 exam years.