Question

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:

 

• A. \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} + 2 \right) \)
• B. \( \frac{\mu_0 I}{2 \pi a} \left( \frac{\pi}{2} + 2 \right) \)
• C. \( \frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} \right) \)
• D. \( \frac{\mu_0 I}{2 \pi a} \left( \frac{3\pi}{2} + 1 \right) \)

Hint:

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.

Answer 1

The Correct Option is C

Let the magnetic fields due to different segments of the wire be \( B_1 \), \( B_2 \), and \( B_3 \). For the arc with radius \( a \) and angle \( \frac{3\pi}{2} \), the magnetic field at the origin is: \[ B_1 = \frac{\mu_0 I}{4\pi a} \] For the straight segment of the wire: \[ B_2 = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \] Since the magnetic field due to the straight segments at the origin is zero: \[ B_3 = 0 \] Thus, the total magnetic field at the origin is: \[ B = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]