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: Use the formula for magnetic field at the center of a circular coil. The magnetic field at the center of a circular coil is given by: \[ B = \frac{\mu_0 N I}{2 R} \] Where: - \( N = 250 \) is the number of turns, - \( I = 20 \, \text{mA} = 20 \times 10^{-3} \, \text{A} \) is the current, - \( R = 0.4 \, \text{m} \) is the radius, - \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space.
Step 2: Substitute the values: \[ B = \frac{(4 \pi \times 10^{-7}) \times 250 \times 20 \times 10^{-3}}{2 \times 0.4} \]
Step 3: Final result: \[ B = 5.25 \times 10^{-5} \, \text{T} \]