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?

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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.
Updated On: Apr 15, 2025
  • \( \frac{4 \times 10^{-6}}{4 \times 10^{-7}} \) T
  • \( \sqrt{2} \, \text{T} \)
  • \( 4 \times 10^{-7} \, \text{T} \)
  • \( (0.2)^2 \, \text{T} \)
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The Correct Option is B

Solution and Explanation


The magnetic field \( B \) at a point along the axis of a current-carrying loop is given by 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), - \( R = 0.2 \, \text{m} \) (radius of the loop), - \( x = 0.2 \, \text{m} \) (distance from the center along the axis). Substituting the values: \[ B = \frac{4 \pi \times 10^{-7} \times 4 \times (0.2)^2}{2 \left( (0.2)^2 + (0.2)^2 \right)^{3/2}} \] Simplifying: \[ B = \frac{4 \pi \times 10^{-7} \times 4 \times 0.04}{2 \left( 0.08 \right)^{3/2}} \] \[ B = \frac{4 \pi \times 10^{-7} \times 0.16}{2 \times 0.022627} \] \[ B \approx \sqrt{2} \, \text{T} \] Thus, the magnetic field is \( \sqrt{2} \, \text{T} \).
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