Question

A circular coil of wire consisting of ' $n$ ' turns each of radius $8\text{ cm}$ carries a current of $0.4\text{ A}$ . The magnitude of the magnetic field at the centre of coil is $3.14 \times 10^{-4}\text{ T}$. The value of ' $n$ ' is [Take $\mu_0 = 12.56 \times 10^{-7}\text{ SI unit}$]}

• A. $1$
• B. $10$
• C. $100$
• D. $1000$

Hint:

For magnetic field at the centre of a circular coil: \[ B=\frac{\mu_0 nI}{2R} \] Always convert radius into metres before substitution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A circular loop's magnetic field at its center is summed over all its turns.
Step 2: Key Formula or Approach:
$B = \frac{\mu_0 n I}{2R}$.
Step 3: Detailed Explanation:
$3.14 \times 10^{-4} = \frac{12.56 \times 10^{-7} \times n \times 0.4}{2 \times 0.08}$.
$3.14 \times 10^{-4} = \frac{5.024 \times 10^{-7} \times n}{0.16} \implies 3.14 \times 10^{-4} = 3.14 \times 10^{-6} \times n \implies n = 100$.
Step 4: Final Answer:
The number of turns is 100.