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. $\frac{{\mu }_{0}j}{4r}$
• B. $\frac{{\mu }_0{i}^2}{4r}$
• C. $\frac{{\mu }_{0}i}{4r^2}$
• 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 $ab$ 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
$B=\Sigma \delta B=\Sigma \frac{{\mu }_0}{4\pi }\frac{i\delta l}{r^2}$
$=\frac{{\mu }_0}{4\pi }\frac{i}{r^2}\Sigma \delta l$
$=\frac{\mu \dot{\alpha }}{4\pi r^2}(\pi r)=\frac{{\mu }_j}{4r}$