question_answer

A straight thin conductor is bent as shown in the adjoining figure. It carries a current i ampere. The radius of the semicircular arc is r metre. The magnetic induction at the centre of semicircular arc is:

A) \[\frac{{{\mu }_{0}}}{4\pi }\frac{i}{r}tesla\]                         

B)  \[\frac{{{\mu }_{0}}}{4\pi }\frac{i}{r}tesla\]

C)  \[\frac{{{\mu }_{0}}}{4\pi }\frac{i}{2r}tesla\]                      

D)  zero

Correct Answer: B

Solution :

                From Biot-Savarts law, the magnetic field at a point distant r from a current element\[\delta l\]is \[\delta B=\frac{{{\mu }_{0}}}{4\pi }\frac{i\delta l\sin \theta }{{{r}^{2}}}\] where\[\theta \]is the angle between the element\[\delta l\]and the line joining the element to that point. For each straight part of wire\[\theta =0\](or\[{{180}^{o}}\]). Hence, for each straight point, we have                 \[{{B}_{1}}={{B}_{2}}=0\] The field at centre 0 due to the semi-circular current loop of radius r is                 \[{{B}_{3}}=\frac{1}{2}\left( \frac{{{\mu }_{0}}i}{2r} \right)=\frac{{{\mu }_{0}}i}{4r}\] Therefore, the field due to the full wire is \[{{B}_{3}}={{B}_{1}}+{{B}_{2}}+{{B}_{3}}=\frac{{{\mu }_{0}}i}{4r}\]tesla