Question

A wire is bent in the form of a circular arc with a straight portion $AB$. Magnetic induction at $O$ when current $I$ is flowing in the wire, is

• A. $\frac{{\mu }_0}{2r}(\pi -\theta +\tan \theta )$
• B. $\frac{{\mu }_{0}I}{2\pi r}(\pi +\theta -\tan \theta )$
• C. $\frac{{\mu }_{0}I}{2\pi r}(\pi -\theta +\tan \theta )$
• D. $\frac{{\mu }_{0}I}{2\pi r}(-\tan \theta +\pi -\theta )$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{2\pi r}(\pi -\theta +\tan \theta )$

$B=\frac{{\mu }_{0}I}{4\pi \left(OC\right)}\left[2\sin \theta \right]$

But $OC=r\cos \theta or{B}_{AB}=\frac{{\mu }_{0}I}{2\pi }\tan \theta$

Magnetic field due to circular portion,

$B_{AB}=\frac{{\mu }_{0}I}{2r}\left(\frac{2\pi -2\theta }{2\pi }\right)=\frac{{\mu }_{0}I}{2r}(\pi -\theta )$

Total magnetic field

$=\frac{{\mu }_{0}I}{2\pi }\tan \theta +\frac{{\mu }_{0}I}{2r}(\pi -\theta )=\frac{{\mu }_{0}I}{2\pi r}[\tan \theta +\pi -\theta ]$