First slide

Magnetic field due to a current carrying circular ring

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

Moderate

A

nB
B

$n^{2} B$
C

2nB
D

$2 n^{2} B$

Solution

Let l be the length of the wire. Magnetic field at the centre of the loop is
$B = \frac{μ_{0} I}{2 R}$
$∴ B = \frac{μ_{0} π I}{l} (∵ l = 2 π R)$ ……….(1)

$B^{'} = \frac{μ_{0} n I}{2 r} = \frac{μ_{0} n I}{2 (\frac{l}{2 n π})}$
$B^{'} = \frac{μ_{0} n^{2} π I}{l}$
From eqns. (i) and (ii), we get
$B^{'} = n^{2} B$

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