Question

A wire has three different sections as shown in figure. The magnitude of the magnetic field produced at the centre '\(O\)' of the semicircle by three sections together is ( \(\mu_0 =\) permiability of free space)

• A. \(\frac{\mu_0 \text{I}}{4\text{R}}\)
• B. \(\frac{\mu_0 \text{I}}{2\text{R}}\)
• C. \(\frac{\mu_0 \text{I}}{4\pi\text{R}}\)
• D. \(\frac{\mu_0 I}{2\pi R}\)

Hint:

If a straight wire segment is in line with the point $O$, ignore it. For arcs, the field is $(\text{angle in radians}/2\pi) \times (\mu_0 I/2R)$. Semicircle $= \pi/2\pi = 1/2$.

Answer 1

The Correct Option is A

Step 1: Concept
According to the Biot-Savart law, a straight wire segment produces zero magnetic field at any point along its axis.

Step 2: Meaning
The two straight sections point directly toward or away from the center $O$, so their contribution is $B_{straight} = 0$.

Step 3: Analysis
Only the semicircular arc contributes to the field at $O$.
Magnetic field of a full circle at center $= \frac{\mu_0 I}{2R}$.
Magnetic field of a semicircle $= \frac{1}{2} \times \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R}$.

Step 4: Conclusion
The total magnetic field is $\frac{\mu_0 I}{4R}$. Final Answer: (A)