Question

As shown in the figure, a very long conducting wire is bent in a semi-circular shape from A to B. The magnetic field at point P for steady current configuration is given by:

• A. \(\frac{\mu_0i}{4R}\) pointed away from the page
• B. \(\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed away from the page
• C. \(\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed into the page
• D. \(\frac{\mu_0i}{4R}\) pointed into the page

Answer 1

The Correct Option is B

To determine the magnetic field at point P due to the semi-circular loop of a conducting wire, we use the Biot-Savart Law. The Biot-Savart Law for a current-carrying wire is given by:

\(dB = \frac{\mu_0}{4\pi} \cdot \frac{i \cdot dl \times \mathbf{r}}{r^3}\)

For a circular arc of radius \(R\) and current \(i\), the magnetic field at the center due to an arc of angle \(\theta\) is:

\(B = \frac{\mu_0i\theta}{4\pi R}\)

Since the arc here is semi-circular, \(\theta = \pi\). Thus, the magnetic field at point P due to the semi-circular part is:

\(B_{\text{semi-circle}} = \frac{\mu_0i\pi}{4\pi R} = \frac{\mu_0i}{4R}\)

This field direction due to the semi-circle will be into the page (using the right-hand rule).

In addition to the arc, there are two straight parts of the wire extending in opposite directions, which contribute zero net magnetic field at point P as they are symmetrical and infinitely long.

Thus, the net magnetic field at point P is:

\(B_{\text{net}} = B_{\text{semi-circle}} - B_{\text{straight parts}}\)

So, considering the directionality:

\(B_{\text{total}} = \frac{\mu_0i}{4R} - \frac{2\mu_0i}{4\pi R} = \frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed away from the page.

Therefore, the correct answer is:

\(\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed away from the page.