question_answer

An infinite straight current carrying conductor is bent into a circle as shown in the figure. If the radius of the circle is R, the magnetic field at the centre of the coil is:

A)  infinite                                

B)  zero

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

D) \[\frac{{{\mu }_{0}}2i}{4\pi R}(\pi +1)\]

Correct Answer: D

Solution :

                The magnitude of magnetic field at 0 due to the straight part of wire is \[{{B}_{1}}=\frac{{{\mu }_{0}}}{2\pi }\frac{i}{r}\] \[{{B}_{1}}\]is perpendicular to the plane of page, directed downwards. The field at the centre O due to current loop of radius r is \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\]                        ...(ii) \[{{B}_{2}}\]is also perpendicular to page, directed upwards. \[\therefore \]Resultant field at O is                 \[{{B}_{1}}+{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\left( \frac{1}{\pi }+1 \right)\]                 \[=\frac{{{\mu }_{0}}2i}{4\pi r}(\pi +1)\] Perpendicular to page directed upwards.