Question

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

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

$B_{AB}=\frac{{\mu }_{0}I}{4\pi \left(OC\right)}\left(2\sin \theta \right)$
But $OC$ = $r\cos \theta$ or $B_{AB}=\frac{{\mu }_{0}I}{2\pi r}\tan \theta$

Magnetic field due to circular portion,
$B_{AB}=\frac{{\mu }_{0}I}{2r}\times \frac{(2\pi -2\theta )}{2\pi }=\frac{{\mu }_{0}I}{2\pi r}(\pi -\theta )$

Total magnetic field = $\frac{{\mu }_{0}I}{2\pi r}\tan \theta +\frac{{\mu }_{0}I}{2\pi r}(\pi -\theta )=\frac{{\mu }_{0}I}{2\pi r}(\pi -\theta +\tan \theta )$