question_answer

Apart of a long wire carrying a current \[i\] is bent into a circle of radius \[r\] as shown in figure. The net magnetic field at the centre \[O\] of the circular loop is

A) \[\frac{{{\mu }_{0}}I}{4r}\]                                         

B) \[\frac{{{\mu }_{0}}I}{2r}\]

C) \[\frac{{{\mu }_{0}}I}{2\pi r}(\pi +1)\]                    

D) \[\frac{{{\mu }_{0}}I}{2\pi r}(\pi -1)\]

Correct Answer: C

Solution :

The magnitude of the magnetic field at point \[O\] due to straight part of wire is \[{{B}_{1}}\]is perpendicular to the plane of the page, directed upwards (right hand palm rule1). The field at the centre 0 due to the current loop of radius \[r\] is                 \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\] \[{{B}_{2}}\]is also perpendicular to the page, directed upwards (right hand screw rule). \[\therefore \]Resultant field at \[O\] is                 \[{{B}_{1}}+{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\left( \frac{1}{\pi }+1 \right)\]                 \[=\frac{{{\mu }_{0}}i}{2\pi r}(\pi +1)\]