Question

A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $n$ turns. If the same current passes in both the cases, the ratio of magnetic fields produced at the centre for one turn to that of $n$ turns is

• A. $1 : n$
• B. $n : 1$
• C. $1 : n^2$
• D. $n^2 : 1$

Hint:

When a wire of fixed length is coiled into $n$ turns, the radius becomes $1/n$ of its original value, and the magnetic field at the center increases by a factor of $n^2$ because $B \propto \frac{n}{r}$.

Answer 1

The Correct Option is C

Step 1: Let $L$ be the total length of the wire. For a coil of one turn ($n_1 = 1$), the radius $r_1$ is related to the length by $L = 2\pi r_1$, which gives: \[ r_1 = \frac{L}{2\pi} \]
Step 2: For a coil of $n$ turns ($n_2 = n$), the radius $r_2$ is related to the length by $L = n(2\pi r_2)$, which gives: \[ r_2 = \frac{L}{2n\pi} \]
Step 3: 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 4: For the first case (single turn): \[ B_1 = \frac{\mu_0 (1) I}{2 r_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L} \]
Step 5: For the second case ($n$ turns): \[ B_2 = \frac{\mu_0 n I}{2 r_2} = \frac{\mu_0 n I}{2(L/2n\pi)} = \frac{\mu_0 \pi n^2 I}{L} \]
Step 6: The ratio of magnetic fields is: \[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 \pi I}{L{\frac{\mu_0 \pi n^2 I}{L = \frac{1}{n^2} \] Therefore, the ratio is $1 : n^2$.