JEE Main · 2004 · 24 Apr 2004 · Offline · Magnetic Effects of Current

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A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

(A)nB
(B) $n^{2} B$ ✓ Correct
(C)2nB
(D) $2 n^{2} B$

Answer: (B)

Solution

A single circular turn of radius

r

produces a central field

B = \frac{μ _{0} i}{2 r}

. Winding the same wire of length

ℓ

into

n

turns makes each turn's radius

r^{'} = \frac{ℓ}{2 π n}

, and with

n

turns the central field is

B^{'} = \frac{n μ _{0} i}{2 r ^{'}} = \frac{n μ _{0} i}{2} \cdot \frac{2 π n}{ℓ} = \frac{n ^{2} μ _{0} iπ}{ℓ}

. Compared with the single turn, the field is

n^{2}

times as large.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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