Question

A circular coil of radius 40 cm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is:

• A. \( 5.25 \times 10^{-5} \) T
• B. \( 2.50 \times 10^{-5} \) T
• C. \( 7.85 \times 10^{-5} \) T
• D. \( 6.20 \times 10^{-5} \) T

Hint:

The magnetic field at the center of a circular coil is directly proportional to the current and the number of turns, and inversely proportional to the radius.

Answer 1

The Correct Option is A

Step 1: Concept
A current-carrying circular coil produces a magnetic field at its center given by:
\( B = \frac{\mu_0 N I}{2R} \)

Step 2: Substitution
Given:
\( N = 250,\ I = 20 \times 10^{-3} \text{ A},\ R = 0.4 \text{ m} \)

\[ B = \frac{(4\pi \times 10^{-7}) \cdot 250 \cdot 0.02}{2 \cdot 0.4} \]

Step 3: Result
\[ B \approx 7.85 \times 10^{-6}\ \text{T} \]

Final Answer:
Thus, the magnetic field at the center is \(7.85 \times 10^{-6}\ \text{T}\).