question_answer

A part 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 \[{{B}_{1}}=\frac{{{\mu }_{0}}i}{2\pi r}\] \[{{B}_{1}}\]is perpendicular to the plane of the page, directed upwards (right hand palm rule 1). The field at the centre O 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 \[{{B}_{1}}+{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\left( \frac{1}{\pi }+1 \right)\] \[=\frac{{{\mu }_{0}}i}{2\pi r}(\pi +1)\]