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

Moving Charges and Magnetism in past papers

89 questions from this chapter have appeared across 27 exam years.