Question

A long wire having a semicircular loop of radius $r$ carries a current $i$ as shown in figure. The magnetic induction at the center $O$ due to entire wire is

• A. $\frac{\mu_{0} j}{4 r}$
• B. $\frac{\mu_{0}i^{2}}{4 r}$
• C. $\frac{\mu_{0}i}{4 r^{2}}$
• D. None of these

Answer 1

Answer: (A)

According to Biot-Savart Law, the magnetic induction at a point to a current carrying element $i \delta l$ is given by
$\delta B =\frac{\mu_{0}}{4 \pi} \frac{i \delta 1 \times r }{r^{3}}$
or $B=\frac{\mu_{0}}{4 \pi} \frac{i \delta l \sin \theta}{r^{2}}$
Directed normal to plane containing $\delta 1$ and $r \theta$ being angle between $\delta l$ and $r$
Magnetic field due to linear portions of wire
If we consider any element of length $\delta 1$ of either linear portion $a b$ or de, the angle between $\delta l$ and $r$ is 0 or $\pi$ therefore,
$\sin \theta=0$
Hence, magnetic induction due to both linear portions of wire is
$\frac{\mu_{0}}{4 \pi} \frac{i \delta l \sin \theta}{r}=0$
Field due to semi-circular are
Now angle between a current element $\delta l$ of semi-circular arc and the radius vector of the element to point $c$ is $\pi / 2$.
Therefore, the magnitude of magnetic induction $B$ at $O$ due to this element is
$\delta B=\frac{\mu_{0}}{4 \pi} \frac{i \delta l \sin \pi / 2}{r^{2}}=\frac{\mu_{j} \delta l}{4 \pi r^{2}}$
Hence, magnetic induction due to whole semi-circular loop is
$B =\Sigma \delta B=\Sigma \frac{\mu_{0}}{4 \pi} \frac{i \delta l}{r^{2}}$
$=\frac{\mu_{0}}{4 \pi} \frac{i}{r^{2}} \Sigma \delta l$
$=\frac{\mu \dot{\alpha}}{4 \pi r^{2}}(\pi r)=\frac{\mu_{j}}{4 r}$