Question

In the following figure, the magnitude of the magnetic field at point 'O' will be

• A. \(\frac{μ_0}{4π}\frac{I}{r}(\frac{2}{π}+2)\)
• B. \(\frac{μ_0}{4π}\frac{I}{r}(\frac{2}{π}-2)\)
• C. \(\frac{μ_0}{4π}\frac{I}{r}(2+\frac{π}{2})\)
• D. \(\frac{μ_0}{4π}\frac{I}{r}(2-\frac{π}{2})\)

Hint:

Add the fields from the two semi-infinite straight wires and the quarter circle arc.

Answer 1

The Correct Option is C

Step 1: Use the Biot-Savart result for a semi-infinite wire:
With the point on the perpendicular through the end of the wire, the field is half that of an infinite wire: $\frac12\cdot\frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi r}$.

Step 2: Add the three parts:
The arc gives $\frac{\mu_0 I}{2r}\cdot\frac14 = \frac{\mu_0 I}{8r} = \frac{\mu_0}{4\pi}\frac Ir\cdot\frac\pi2$.
All three contributions have the same direction, so they add: $\frac{\mu_0}{4\pi}\frac Ir\left(2 + \frac\pi2\right)$.

Final Answer:
The field is $\frac{\mu_0}{4\pi}\frac{I}{r}\left(2 + \frac{\pi}{2}\right)$, option (C). \[ \boxed{\frac{\mu_0}{4\pi}\frac{I}{r}\left(2+\frac{\pi}{2}\right)} \]