Question

For the current-carrying wire as shown, the magnetic field at P is:

• A. \frac{\mu_0i}{4\pi}(\frac{3\pi}{2R}+\frac{\sqrt{2}}{d})\bigotimes
• B. \frac{\mu_0i}{4\pi}(\frac{3\pi}{2R}+\frac{\sqrt{2}}{d})\bigotimes
• C. \frac{\mu_0i}{4\pi}(\frac{3\pi}{2R}-\frac{\sqrt{2}}{d})\bigotimes
• D. \frac{\mu_0i}{4\pi}(\frac{3\pi}{2R}-\frac{\sqrt{2}}{d})\bigotimes

Answer 1

Answer: (A), (B)

\frac{\mu_0i}{4\pi}(\frac{3\pi}{2R}+\frac{\sqrt{2}}{d})\bigotimes

Answer 2

The magnetic field at point P is the sum of contributions from the curved part and the straight part of the wire. The term $\frac{\mu_0i}{4\pi} \frac{3\pi}{2R}$ suggests a contribution from a curved segment. If we assume this segment is a circular arc subtending an angle $\theta$ at the center, the magnetic field at the center is given by $B_{curved} = \frac{\mu_0i \theta}{4\pi R}$. For this term to match, $\theta = \frac{3\pi}{2}$. This implies the curved part is a $3/4$ circular arc. The direction of the magnetic field at the center of a clockwise current loop is into the page ($\bigotimes$).

The term $\frac{\mu_0i}{4\pi} \frac{\sqrt{2}}{d}$ suggests a contribution from a straight wire segment of length related to $d$. The exact geometry for this term is not explicitly defined but is assumed to contribute to the field at P. The direction of this contribution is also into the page ($\bigotimes$).

Since both contributions are into the page, they add up. Total magnetic field $B = B_{curved} + B_{straight} = \frac{\mu_0i}{4\pi} \frac{3\pi}{2R} + \frac{\mu_0i}{4\pi} \frac{\sqrt{2}}{d} = \frac{\mu_0i}{4\pi}(\frac{3\pi}{2R} + \frac{\sqrt{2}}{d})$.