A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane and the other in $x-z$ plane. If the current in the loop is $i$. The resultant magnetic field due to the two semicircular parts at their common centre is

Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength $ B = 0.5 \, \text{T} $. If the current in the wire is $ I = 2 \, \text{A} $ and the wire makes an angle of $ 30^\circ $ with the magnetic field, find the force per unit length on the wire.

Answer 1

The problem involves calculating the resultant magnetic field at the center of a current loop consisting of two semicircular parts, each of radius $R$. One semicircle lies in the $x-y$ plane and the other in the $x-z$ plane.

The magnetic field at the center of a complete circular loop carrying current $i$ is given by the formula:

$$B = \frac{\mu_0 i}{2 R}$$

Since each semicircular section is half of a full circle, the magnetic field due to each semicircle at the center will be half of that produced by a complete circular loop. Therefore, the magnetic field produced by each semicircular part at the center is:

$$B_{\text{semicircle}} = \frac{1}{2} \times \frac{\mu_0 i}{2 R} = \frac{\mu_0 i}{4 R}$$

Both magnetic fields produced by the semicircles will be perpendicular to each other because they lie in two different planes ($xy$ and $xz$ planes). Therefore, the resultant magnetic field $B_{\text{resultant}}$ can be found using the Pythagorean theorem:

$$B_{\text{resultant}} = \sqrt{B_{\text{semicircle, }xy}^2 + B_{\text{semicircle, }xz}^2}$$

Substituting the expressions for the magnetic fields from the semicircles:

$$B_{\text{resultant}} = \sqrt{\left(\frac{\mu_0 i}{4 R}\right)^2 + \left(\frac{\mu_0 i}{4 R}\right)^2}$$

$$B_{\text{resultant}} = \sqrt{2 \times \left(\frac{\mu_0 i}{4 R}\right)^2}$$

$$B_{\text{resultant}} = \frac{\mu_0 i}{4 R} \times \sqrt{2} = \frac{\mu_0 i}{4 R} \times \frac{\sqrt{2}}{2} \times 2$$

$$B_{\text{resultant}} = \frac{\mu_0 i}{2 \sqrt{2} R}$$

Thus, the resultant magnetic field at the center due to two perpendicular semicircular parts of the loop is $\frac{\mu_0 i}{2\sqrt{2} R}$.

Answer 2