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 . It is then bent into a circular coil of turns. The magnetic field at the centre of this coil of turns will be

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Solution

Q.
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 [2016]

1 $n B$

2 $n^{2} B$

3 $2 n B$

4 $2 n^{2} B$

Ans.
(2)

Let $l$ be the length of the wire. Magnetic field at the centre of the loop is

$B = \frac{μ_{0} I}{2 R} ∴ B = \frac{μ_{0} π I}{l} (∵ l = 2 π R)$ ...(i)

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

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