Question

An infinitely long straight conductor is bent into shape as shown in figure. It carries a current I A. and the radius of circular loop is r metre. Then the magnetic induction at the centre of the circular loop is:

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

Answer 1

The correct option is D $\frac{{\mu }_{0}i}{2\pi r}(\pi -1)$

magnetic field due to circular path will be along -ve z axis.

$B_{circle}=\frac{{\mu }_{0}I}{2R}(-\hat{k})$

magnetic field due to straight wire will be long +ve z axis

$B_{straight}=\frac{{\mu }_{0}I}{2\pi R}\left(\hat{k}\right)$

$\therefore \overrightarrow{B_{next}}=-\frac{{\mu }_{0}I}{2R}\left(\hat{k}\right)+\frac{{\mu }_{0}I}{2\pi R}\left(\hat{k}\right)=\frac{{\mu }_{0}I}{2R}(-1+\frac{1}{\pi })\hat{k}$

$\therefore \left|\overrightarrow{B_{next}}\right|=\frac{{\mu }_{0}i}{2\pi R}(\pi -1)$