Question

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

• A. $\frac{{\mu }_{0}I}{4\pi r}(\frac{3\pi }{2}+1)$
• B. $\frac{{\mu }_{0}I}{4\pi r}(\frac{3\pi }{2}-1)$
• C. $\frac{{\mu }_{0}I}{4\pi r}(\frac{\pi }{2}+1)$
• D. $\frac{{\mu }_{0}I}{4\pi r}(\frac{\pi }{2}-1)$

Answer 1

Answer: (A)

$B_A=0$

$B_B=\frac{{\mu }_0}{4\pi }\frac{(2\pi -\pi /2)I}{r}\otimes$
$=\frac{{\mu }_0}{4\pi }\frac{3\pi I}{2r}$
$B_C=\frac{{\mu }_{0}I}{4\pi r}\otimes$
So, net magnetic field at the centre
$=B_A+B_B+B_C$
$=0+\frac{{\mu }_0}{4\pi }\frac{3\pi I}{2r}+\frac{{\mu }_{0}I}{4\pi r}=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\frac{3\pi }{2}+1\right)$