Question

A long wire bent as shown in figure carries current $I$. If the radius of the semicircular portion is $a$, the magnetic field at center $C$ is:

• A. $\frac{{\mu }_{0}I}{4a}$
• B. $\frac{{\mu }_{0}I}{4\pi a}\sqrt{{\pi }^2+4}$
• C. $\frac{{\mu }_{0}I}{4a}+\frac{{\mu }_{0}I}{4\pi a}$
• D. $\frac{{\mu }_{0}I}{4\pi a}\sqrt{{\pi }^2-4}$

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{4\pi a}\sqrt{{\pi }^2+4}$

Due to semicircular part: $B_1=\frac{1}{2}\left(\frac{{\mu }_{0}I}{2a}\right)$
$B_2=2\times (\frac{1}{2}\times (\frac{{\mu }_{0}I}{2\pi a}\left)\right)$
The factor of 2 outside is due to two semi-infinite wires.
Hence, resultant field,
$B=\sqrt{{B_1}^2+{B_2}^2}=\frac{{\mu }_{0}I}{2a}\sqrt{\left(\right(\frac{1}{2}{)}^2+\left(\frac{1}{\pi }{)}^2\right)}=\frac{{\mu }_{0}I}{4\pi a}\sqrt{4+{\pi }^2}$