Question Detail
Question
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 for same current will be
Correct Answer
Option B is the correct answer.
Detailed Explanation
The magnetic field (B) at the center of a circular coil with N turns and radius R, carrying a current I, is given by B = (μ0NI) / (2R).Initial state:N1 = 1 turn, radius R1.B1 = (μ0 × 1 × I) / (2R1) = BFinal state:N2 = n turns, radius R2.B2 = (μ0nI) / (2R2)The total length of the wire remains constant. Let L be the length.L = N1(2πR1) = 1 × 2πR1 = 2πR1L = N2(2πR2) = n × 2πR2From this, 2πR1 = n × 2πR2, which implies R1 = nR2, or R2 = R1/n.Substitute R2 into the expression for B2:B2 = (μ0nI) / (2 × (R1/n)) = (μ0n2I) / (2R1)Comparing B2 with B1:B2 = n2 × (μ0I / (2R1)) = n2BThus, the magnetic field at the center will be n2B.
Hint
Remember the formula for the magnetic field at the center of a coil. The total length of the wire remains constant, which implies a relationship between the initial and final radii of the coil.