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 int - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Let the length of the wire be L .Case 1: Single turn (n = 1). L = 2 π R ⟹ R = \frac{L}{2 π} .Magnetic field B = \frac{μ _{0} I}{2 R} = \frac{μ _{0} I π}{L} .Case 2: n turns. L = n (2 π r) ⟹ r = \frac{L}{2 πn} = \frac{R}{n} .Magnetic field B^{'} = \frac{μ _{0} n I}{2 r} = \frac{μ _{0} n I}{2 ( R / n )} = n^{2} (\frac{μ _{0} I}{2 R}) = n^{2} B .

Try it yourself before checking the explanation above.

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:

n B

n^{2} B

2 n B

2 n^{2} B

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