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 is passed in both the cases, the ratio of magnetic field 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:

The same wire length is used, so the radius becomes R/n when there are n turns.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The field at the centre of a circular coil of N turns and radius r is \(B = \dfrac{\mu_0NI}{2r}\).

Step 2: Radius change
The same wire length \(2\pi R\) is used. For n turns, \(n \cdot 2\pi r = 2\pi R\), so \(r = R/n\).

Step 3: Ratio
\[ B_1 = \frac{\mu_0I}{2R},\qquad B_n = \frac{\mu_0 n I}{2(R/n)} = \frac{\mu_0 n^2 I}{2R} \]
\[ \frac{B_1}{B_n} = \frac{1}{n^2} \]
So the ratio is \(1 : n^2\). Option (A) forgets that the radius also shrinks.

Final Answer:
The ratio is \(1 : n^2\), option (C). \[ \boxed{1 : n^2} \]