2016 Phase II Moving Charges and MagnetismMagnetic Field due to a Circular Coil Easy Question 7 of 15 in Magnetic Field due to a Circular Coil

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be:

A 2nB
B $2n^2B$
C nB
D $n^2B$

The answer is D.

Solution

n turns and radius R/n give $n^2$ times the field.

Since $l = 2\pi R = n(2\pi r) \Rightarrow r = \frac{R}{n}$
For one turn: $B = \frac{\mu_0i}{2R}$
For n turns: $B' = \frac{\mu_0ni}{2r} = \frac{\mu_0n^2i}{2R} = n^2B$

Concept coil-n-turns-field

Magnetic Field due to a Circular Coil in past papers

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