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: The magnetic field at the centre of a circular coil of \( N \) turns and radius \( r \) carrying current \( I \) is \[ B = \frac{\mu_0 N I}{2r}. \]

Step 2 (single loop): One turn uses the whole wire, so \( 2\pi r_1 = L \Rightarrow r_1 = \dfrac{L}{2\pi} \). With \( N_1 = 1 \), \[ B_1 = \frac{\mu_0 (1) I}{2r_1} = \frac{\mu_0 I}{2\,(L/2\pi)} = \frac{\mu_0 \pi I}{L}. \]

Step 3 (three loops): Three turns use the same wire, so \( 3(2\pi r_2) = L \Rightarrow r_2 = \dfrac{L}{6\pi} \). With \( N_2 = 3 \), \[ B_2 = \frac{\mu_0 (3) I}{2r_2} = \frac{3\mu_0 I}{2\,(L/6\pi)} = \frac{3\mu_0 I \cdot 6\pi}{2L} = \frac{9\mu_0 \pi I}{L}. \]

Step 4: Take the ratio \[ \frac{B_1}{B_2} = \frac{\mu_0 \pi I / L}{9\mu_0 \pi I / L} = \frac{1}{9}. \] So \( B_1 : B_2 = 1:9 \), option 3.

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