2002 AIPMT Moving Charges and MagnetismMagnetic Field due to a Circular Coil Medium Question 15 of 15 in Magnetic Field due to a Circular Coil

The magnetic field of given length of wire for single turn coil at its centre is 'B' then its value for two turns coil for the same wire is : -

A $\dfrac{B}{4}$
B $\dfrac{B}{2}$
C 4B
D 2B

The answer is C.

Solution

For a single turn of radius R: $B_1 = B = \dfrac{\mu_0I}{2R}$
The same wire is wound into 2 turns, so the length stays the same: $2 \times 2\pi r = 2\pi R \Rightarrow r = \dfrac{R}{2}$

For N turns: $B_2 = \dfrac{\mu_0NI}{2r} = \dfrac{\mu_0(2I)}{2(R/2)} = 4 \times \dfrac{\mu_0I}{2R}$
$B_2 = 4B$
In general, rewinding into n turns gives $n^2B$.

Concept Rewinding the same wire into more turns

Magnetic Field due to a Circular Coil in past papers

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