Question

A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is B. If it is bent into a circular loop of radius 'r' having n turns, the magnetic field at the centre of the coil for same current is

• A. \(\frac{\text{B}}{\text{n}^{2}}\)
• B. \(\frac{\text{B}}{\text{n}}\)
• C. \(\text{n}^{2}\text{B}\)
• D. \(\text{nB}\)

Hint:

The wire length is fixed, so n turns make each turn's radius n times smaller.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The field at the centre of a coil is \(B=\dfrac{\mu_0nI}{2r}\) for \(n\) turns of radius \(r\).

Step 2: Single turn
For one turn the radius is \(R=\dfrac{L}{2\pi}\), and \(B=\dfrac{\mu_0I}{2R}\).

Step 3: n turns from the same wire
Each turn has radius \(r=\dfrac{L}{2\pi n}=\dfrac{R}{n}\).

Step 4: New field
\(B'=\dfrac{\mu_0nI}{2(R/n)}=n^2\cdot\dfrac{\mu_0I}{2R}=n^2B\)

Final Answer:
The new field is \(n^2B\). \[ \boxed{\text{(C) }\text{n}^{2}\text{B}} \]