Question

What is the magnetic induction of the field at the point O in a current I carrying wire that has the shape as shown in the figure? The radius of the curved part of the wire is R, the linear parts are assumed to be very long.

• A. $\frac{\mu_{0}I}{4\pi R}[2+\pi]$
• B. $\frac{\mu_{0}I}{4\pi R}[2+\frac{3\pi}{2}]$
• C. $\frac{\mu_{0}I}{4\pi R}[1+\frac{\pi}{2}]$
• D. $\frac{\mu_{0}I}{4\pi R}[1+\frac{3\pi}{2}]$

Hint:

Field from an arc at its center is $B = \frac{\mu_0 I \theta}{4\pi R}$ where $\theta$ is the angle in radians.

Answer 1

The Correct Option is D

Step 1: Concept
The total field at O is the sum of the fields from the straight wires and the circular arc.
Step 2: Analysis
- Field from the horizontal semi-infinite wire: $B_{1} = \frac{\mu_{0}I}{4\pi R}$. - Field from the 3/4 circular arc: $B_{2} = \frac{\mu_{0}I}{2R} \times \frac{3}{4} = \frac{3\mu_{0}I}{8R}$. - The other linear wire passes through O, so its field is zero.
Step 3: Calculation
$B = B_{1} + B_{2} = \frac{\mu_{0}I}{4\pi R} + \frac{3\mu_{0}I}{8R} = \frac{\mu_{0}I}{4\pi R}[1 + \frac{3\pi}{2}]$.
Step 4: Conclusion
Hence, the magnetic induction is $\frac{\mu_{0}I}{4\pi R}[1+\frac{3\pi}{2}]$.
Final Answer:(D)