Question

Current $'I'$ is flowing in a conductor shaped as shown in the figure. The radius of the curved part is $'r'$ and the length of straight portion is very large. The value of the magnetic field at the centre $O$ will be

• A. $\frac {\mu_0I}{4\pi r} ( \frac {3 \pi}{2}+1) $
• B. $\frac {\mu_0I}{4\pi r} ( \frac {3 \pi}{2}-1) $
• C. $\frac {\mu_0I}{4\pi r} ( \frac { \pi}{2}+1) $
• D. $\frac {\mu_0I}{4\pi r} ( \frac { \pi}{2}-1) $

Answer 1

Answer: (A)

$B_{A}=0$

$ B_{B} =\frac{\mu_{0}}{4 \pi} \frac{(2 \pi-\pi / 2) I}{r} \otimes $
$=\frac{\mu_{0}}{4 \pi} \frac{3 \pi I}{2 r} $
$ B_{C} =\frac{\mu_{0} I}{4 \pi r} \otimes$
So, net magnetic field at the centre
$=B_{A}+B_{B}+B_{C}$
$=0+\frac{\mu_{0}}{4 \pi} \frac{3 \pi I}{2 r}+\frac{\mu_{0} I}{4 \pi r}=\frac{\mu_{0}}{4 \pi} \frac{I}{r}\left(\frac{3 \pi}{2}+1\right)$