Question

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. $nB$
• B. $n^2 B$
• C. $2nB$
• D. $2n^2 B$

Answer 1

Answer: (B)

$n^2 B$

Answer 2

One turn loop
$B = \frac{\mu_{0} i}{2R}$
n turn loop
$r = \frac{R}{n}$
$B' = \frac{\mu_{0} ni}{2 \frac{R}{n}}$
$B' = \left(\frac{\mu_{0}ni}{2 R}\right) n^{2}$
$B' = n^{2}B$