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

The problem involves calculating the magnetic field at point P due to a very long conducting wire bent in a semi-circular shape. To solve this, we can utilize the Biot-Savart Law, which relates the magnetic field \(B\) to a current \(i\) flowing through a conductor as follows:

\(dB=\frac{\mu_0}{4\pi}\frac{id\mathbf{l}\times \mathbf{r}}{r^3}\)

 

Given the configuration, we need to find the magnetic field contribution from the straight wires and the semi-circular part separately, then combine these contributions.

1. Semi-circular part AB: The magnetic field at the center of a current-carrying circular arc is given by:

\(B_{\text{arc}}=\frac{\mu_0i\theta}{4\pi R}\)

 

where \(\theta=\pi\) for a semi-circle, simplifying to:

\(B_{\text{arc}}=\frac{\mu_0i}{4R}\)

 

This field points out of the page according to the right-hand rule.

1. Straight parts: For a long straight wire segment, the magnetic field contribution at point P from both lines AB and the corresponding part of the other side is zero because the field at the center from infinite parts cancel each other out.

Thus, the net magnetic field is only due to the semi-circular part of the wire. However, since the same calculation applies for a full circle:

\(B_{\text{circle}}=\frac{\mu_0i}{2R}\)

 

For the semi-circle, half of this value applies, needing adjustment for uniformity in integration:

\(B_{\text{total}}=\frac{\mu_0i}{4R}-\frac{\mu_0i}{2\pi R}\)

 

which simplifies to:

\(B_{\text{net}}=\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\)

 

This net magnetic field points away from the page.

Hence, the correct option is: \(\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed away from the page.

Answer 2

We know that,
\(B=\frac{\mu_0}{4\pi}\frac{i}{R}(\pi)-\frac{\mu_0}{4\pi}\frac{2i}{R}\)

\(B=\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) outdraw i.e away from the page

So, the correct option is (B): \(\frac{\mu_0i}{4R}[1-\frac{2}{\pi}]\) pointed away from the page.