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 ' $\text{B}$ '. If it is bent into a circular loop of radius ' $\text{r}_1$ ' having ' $\text{n}$ ' turns, the magnetic field at the centre of the coil for same current is

• A. $\frac{B}{n^2}$
• B. $\frac{B}{n}$
• C. $n^2 B$
• D. nB

Hint:

More turns = smaller radius + more wire in the same spot. Both factors contribute, leading to the $n^2$ factor.

Answer 1

The Correct Option is C

Step 1: Concept
Magnetic field at the center of a circular coil of $n$ turns is $B = \frac{\mu_0 n I}{2r}$.

Step 2: Meaning
Length of wire is constant: $L = 1 \times 2\pi R = n \times 2\pi r \implies r = R/n$.

Step 3: Analysis
New field $B' = \frac{\mu_0 n I}{2(R/n)} = n \times n \times \frac{\mu_0 I}{2R} = n^2 B$.

Step 4: Conclusion
The magnetic field becomes $n^2 B$. Final Answer: (C)