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:
An infinite wire has a circular bend of radius

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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
  • \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} + 2 \right) \)
  • \( \frac{\mu_0 I}{2 \pi a} \left( \frac{\pi}{2} + 2 \right) \)
  • \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} \right) \)
  • \( \frac{\mu_0 I}{2 \pi a} \left( \frac{3\pi}{2} + 1 \right) \)
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The Correct Option is C

Solution and Explanation

Step 1: Magnetic Field from the Arc

For an arc of radius \( a \) spanning an angle of \( \frac{3\pi}{2} \), the magnetic field at the origin is given by: \[ B_1 = \frac{\mu_0 I}{4\pi a} \]

Step 2: Magnetic Field from the Straight Segment

For the straight segment of the wire, the magnetic field at the origin is calculated as: \[ B_2 = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]

Step 3: Magnetic Field from Other Segments

The magnetic field generated by the straight segments at the origin is zero: \[ B_3 = 0 \]

Step 4: Total Magnetic Field Calculation

The total magnetic field at the origin is the sum of the fields from all segments: \[ B = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]

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