Question

A wire having a semi circular loop of radius $r$ carries a current $i$ as shown in figure. The magnetic field at the center $\text{C}$ due to entire wire is

• A. $\frac{3{\mu }_{0}i}{4r}$
• B. $\frac{{\mu }_{0}i}{2r}$
• C. $\frac{{\mu }_{0}i}{4r}$
• D. $\frac{{\mu }_{0}i}{8r}$

Answer 1

The correct option is C $\frac{{\mu }_{0}i}{4r}$

The magnetic field at $\text{C}$ due to the straight portion of wire will be zero.

$\because \overrightarrow{dl}\times \overrightarrow{r}=0$

For the semicircular segment at any position on wire we find that the current elemeny $\left(\overrightarrow{dl}\right)$ is perpendicular to position vector of point $\text{C}$ $\left(\overrightarrow{r}\right)$.

Thus, using biot-savarts law,

$\text{dB}=\frac{{\mu }_{0}i}{4\pi }\frac{\left(dl\right)\left(r\right)sin{90}^{\circ }}{r^3}=\frac{{\mu }_{0}i}{4\pi r^2}dl$

Now, the net magnetic field at center $\text{C}$ is due to the sum of all the small current elements on the semicircular wire.

$\Rightarrow B=\int dB$

or, $B=\frac{{\mu }_{0}i}{4\pi r^2}\int dl$

Here, $\int dl$ $=$ perimeter of semicircle $=\pi r$

$\therefore B=\frac{{\mu }_{0}i}{4\pi r^2}\times \pi r=\frac{{\mu }_{0}i}{4r}$

$\begin{array}{c}\text{Why this question?}\\ \text{Using Biot-savart law we can find the magnetic field due to any configuration of current carrying wire segments with the help of calculus.}\end{array}$