Question

Find the magnetic field at the point $P$ in figure. The curved portion is a semicircle connected to two long straight wires

• A. $\frac{\mu_0 i}{2 r}\left(1+\frac{1}{\pi}\right)$
• B. $\frac{\mu_0 i}{2 r}\left(1+\frac{2}{\pi}\right)$
• C. $\frac{\mu_0 i}{2 r}\left(\frac{1}{2}+\frac{1}{\pi}\right)$
• D. $\frac{\mu_0 i}{2 r}\left(\frac{1}{2}+\frac{1}{2 \pi}\right)$

Answer 1

The Correct Option is D

To find the magnetic field at the point \( P \), let's consider each part of the conductor separately and apply the Biot-Savart Law and Ampere's Law where appropriate.

1. The semicircular part of the wire contributes to the magnetic field at point \( P \). The magnetic field due to a semicircular arc of radius \( r \) carrying current \( i \) is given by: \(B_{\text{semi}} = \frac{\mu_0 i}{4r}\) (since it's a half-circle).
2. Next, consider the long straight wires. A long straight wire carrying current \( i \) produces a magnetic field at a point at distance \( r \) perpendicular to the wire as: \(B_{\text{straight}} = \frac{\mu_0 i}{2 \pi r}\).
3. In the configuration shown, there are two contributions from the two straight wires at point \( P \), both at the same distance \( r \). However, the directions depend on the right-hand rule.
4. Add the magnetic fields vectorially. Using the right-hand rule:
   • For the semicircle, the field is directed into or out of the page depending on the current direction.
   • For the straight wires, assume they contribute perpendicularly to the point \( P \).
5. The total magnetic field at point \( P \) is: \(B_{\text{total}} = B_{\text{semi}} + 2 \times B_{\text{straight}}\), since both straight wires add equally:
6. Substituting the values: \(B_{\text{total}} = \frac{\mu_0 i}{4r} + 2 \cdot \frac{\mu_0 i}{2 \pi r} = \frac{\mu_0 i}{4r} + \frac{\mu_0 i}{\pi r}\).
7. Put under common denominator: \(B = \frac{\mu_0 i}{2r} \left( \frac{1}{2} + \frac{1}{2 \pi} \right)\).

Thus, the correct answer is: \(\frac{\mu_0 i}{2 r}\left(\frac{1}{2}+\frac{1}{2 \pi}\right)\).