2022 Phase 2 Moving Charges and MagnetismMagnetic Field due to a Circular Coil Medium Question 5 of 15 in Magnetic Field due to a Circular Coil

The magnetic field on the axis of a circular loop of radius 100 cm carrying current $I=\sqrt{2}$ A, at point 1 m away from the centre of the loop is given by:

A $6.28\times10^{-4}$ T
B $3.14\times10^{-7}$ T
C $6.28\times10^{-7}$ T
D $3.14\times10^{-4}$ T

The answer is B.

Solution

$B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}=\frac{4\pi\times10^{-7}\times\sqrt{2}\times1^2}{2(1^2+1^2)^{3/2}}=\frac{4\pi\times10^{-7}\times\sqrt{2}}{2\times2\sqrt{2}}=\pi\times10^{-7}\approx3.14\times10^{-7}$ T.

Concept Magnetic field on the axis of a circular loop

Magnetic Field due to a Circular Coil in past papers

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