Question

Two wires PQ and QR carry equal currents I as shown in figure. One end of both the wires extends to infinity \(\angle PQR = \theta\). The magnitude of the magnetic field at O on the bisector of these two wires at a distance r from point Q is:

• A. \( \frac{\mu_0 I}{4\pi r} \sin\frac{\theta}{2} \)
• B. \( \frac{\mu_0 I}{4\pi r} \cot\frac{\theta}{2} \)
• C. \( \frac{\mu_0 I}{4\pi r} \tan\frac{\theta}{2} \)
• D. \( \frac{\mu_0 I}{2\pi r} \frac{1+\cos(\theta/2)}{\sin(\theta/2)} \)

Hint:

For semi-infinite wires, always use angle-based formula and resolve along symmetry axis.

Answer 1

The Correct Option is D

Concept: Magnetic field due to a semi-infinite straight wire: \[ B = \frac{\mu_0 I}{4\pi r} (\sin\theta_1 + \sin\theta_2) \]

Step 1: Field due to each wire.
Each wire is semi-infinite, so for point O: \[ B_1 = \frac{\mu_0 I}{4\pi r} (1 + \sin\frac{\theta}{2}) \] Similarly for second wire: \[ B_2 = \frac{\mu_0 I}{4\pi r} (1 + \sin\frac{\theta}{2}) \]

Step 2: Direction of fields.
Using right-hand rule, fields due to both wires are symmetric. Resolve along bisector direction.

Step 3: Resultant field.
\[ B = 2B_1 \cos\frac{\theta}{2} \] \[ = 2 \cdot \frac{\mu_0 I}{4\pi r} (1 + \sin\frac{\theta}{2}) \cos\frac{\theta}{2} \] \[ = \frac{\mu_0 I}{2\pi r} \cdot \frac{(1+\cos(\theta/2))}{\sin(\theta/2)} \]