Question:medium

An infinite wire has a circular bend of radius \( a \), and carrying a current \( I \) as shown in the figure. The magnitude of the magnetic field at the origin \( O \) of the arc is given by:

 

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For magnetic fields due to current-carrying wires, always break down the problem into simpler segments (arc, straight line) and use Biot-Savart's law to compute the contribution from each part.
Updated On: Feb 2, 2026
  • \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} + 2 \right) \)
  • \( \frac{\mu_0 I}{2 \pi a} \left( \frac{\pi}{2} + 2 \right) \)
  • \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} \right) \)
  • \( \frac{\mu_0 I}{2 \pi a} \left( \frac{3\pi}{2} + 1 \right) \)
Show Solution

The Correct Option is C

Solution and Explanation

The magnetic fields from individual wire segments are denoted as \( B_1 \), \( B_2 \), and \( B_3 \).The magnetic field at the origin from the arc segment with radius \( a \) and angle \( \frac{3\pi}{2} \) is given by:\[B_1 = \frac{\mu_0 I}{4\pi a}\]The magnetic field from the straight wire segment is:\[B_2 = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right)\]The magnetic field from the straight segments at the origin is zero:\[B_3 = 0\]Consequently, the total magnetic field at the origin is:\[B = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right)\]
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