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: Combine the two effects
Doubling the turns doubles the field. Halving the radius doubles it again.

Step 2: Multiply
$2\times2=4$, so the field is $4B$, option (A).

Final Answer:
The new field is $4B$, option (A). \[ \boxed{4B} \]