Question

A coil of one loop is formed by a wire of length \( L \), and thereafter a coil of 3 loops is formed by the same wire. If the current remains the same in both cases, then the ratio of the magnetic fields at the centre will be:

• A. \( 1:3 \)
• B. \( 3:1 \)
• C. \( 1:9 \)
• D. \( 9:1 \)

Hint:

For a fixed wire length, \( B \propto N^2 \) because more turns shrink the radius; compare \( N=1 \) with \( N=3 \).

Answer 1

The Correct Option is C

Step 1: For a coil made from a wire of fixed length, express the field only in terms of the number of turns \(N\). If \(N\) turns share a wire of length \(L\), each turn has circumference \(L/N\), so the radius is \(r = \dfrac{L}{2\pi N}\).

Step 2: Substitute into \(B = \dfrac{\mu_0 N I}{2r}\): \(B = \dfrac{\mu_0 N I}{2}\cdot\dfrac{2\pi N}{L} = \dfrac{\mu_0 \pi I\, N^2}{L}\). Hence \(B \propto N^2\) for a fixed wire length and current.

Step 3: Compare \(N_1 = 1\) and \(N_2 = 3\): \(\dfrac{B_1}{B_2} = \dfrac{1^2}{3^2} = \dfrac{1}{9}\).

Step 4: Therefore the fields are in the ratio \(1:9\).

\[\boxed{B_1 : B_2 = 1 : 9}\]