Question:

A current of i amp flows in the loop having circular arc r metre, subtending an angle  as shown in the figure. The magnetic field at centre O of circle is

Updated On: Feb 15, 2025
  • (A) μ0i4πrθ
  • (B) 2μ0i4πr2sinθ
  • (C) 2μ0i2rsinθ
  • (D) μ0i2rsinθ
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The Correct Option is A

Approach Solution - 1

Explanation:
Magnetic field due to the current carrying arc at the centre of curvature B=μ0i4πrθ
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Approach Solution -2

Biot-Savart Law

Biot-Savart law is used to determine the strength of the magnetic field at any point due to a current-carrying conductor.

Consider a very small current element of length dl of a conductor carrying current I, the strength of magnetic field dB at distance r from the element is found to be

\[dB=\frac{\mu_0}{4\pi}\frac{I\,dl\,sin\theta}{r^2}\]

Where

  • µ0 is absolute permeability
  • θ is the angle between dl and r

Using Biot-Savart law, the magnetic field at the center of a current carrying loop of radius R is given by

\[B=\frac{\mu_0}{4\pi}\frac{2\pi I}{R}\]

This equation can be rewritten as

\[B=\frac{\mu_0}{4\pi}\frac{I}{R}(2\pi)\]

Where 2π is the angle subtended by the circular at its center.

Thus, if θ will be the angle subtended by an arc of radius R, its magnetic field at the center is given by

\[B=\frac{\mu_0}{4\pi}\frac{I}{R}(θ)\]
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