Question

A circular loop of radius 0.2 m carries a current of 4 A. What is the magnetic field at a point on the axis of the loop at a distance 0.2 m from the center?

• A. \( \frac{4 \times 10^{-6}}{4 \times 10^{-7}} \) T
• B. \( \sqrt{2} \, \text{T} \)
• C. \( 4 \times 10^{-7} \, \text{T} \)
• D. \( (0.2)^2 \, \text{T} \)

Hint:

Use the formula for the magnetic field due to a current-carrying loop along its axis. The field is stronger at points closer to the loop.

Answer 1

The Correct Option is B

The magnetic field \( B \) on the axis of a current-carrying loop is calculated using the formula: \[ B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \] Where: - \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space) - \( I = 4 \, \text{A} \) (current) - \( R = 0.2 \, \text{m} \) (loop radius) - \( x = 0.2 \, \text{m} \) (axial distance from the center) Substituting the given values: \[ B = \frac{4 \pi \times 10^{-7} \times 4 \times (0.2)^2}{2 \left( (0.2)^2 + (0.2)^2 \right)^{3/2}} \] Simplification leads to: \[ B = \frac{4 \pi \times 10^{-7} \times 4 \times 0.04}{2 \left( 0.08 \right)^{3/2}} \] \[ B = \frac{4 \pi \times 10^{-7} \times 0.16}{2 \times 0.022627} \] \[ B \approx \sqrt{2} \, \text{T} \] The resulting magnetic field is approximately \( \sqrt{2} \, \text{T} \).