Question

The magnetic field at the center of circular coil carrying current '\(I\)' for a single turn of a given length of wire is '\(B\)'. The same wire is bent in a circular coil having two turns. When the same current passes through it, the value of magnetic field becomes

• A. \(4B\)
• B. \(2B\)
• C. \(\frac{B}{2}\)
• D. \(\frac{B}{4}\)

Hint:

Same wire length: with \(N\) turns the radius becomes \(r/N\), so \(B\propto N^2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Field at the centre of a coil is \(B=\dfrac{\mu_0NI}{2r}\).

Step 2: Key Formula or Approach
Wire length \(\ell=2\pi rN\) is fixed, so \(r\propto\dfrac1N\).

Step 3: Detailed Explanation
One turn: \(B=\dfrac{\mu_0I}{2r}\).
Two turns: radius \(=\dfrac r2\), so \(B'=\dfrac{\mu_0(2)I}{2(r/2)}=\dfrac{4\mu_0I}{2r}=4B\).

Final Answer:
The field becomes \(4B\), option (A). \[ \boxed{4B\ \text{(A)}} \]