question_answer

A long current carrying wire is bent as shown in figure. The magnetic field induction at 0 is :             

A)  zero                     

B)  \[\frac{2\sqrt{2}{{\mu }_{0}}i}{\pi L}\]

C)  \[\frac{{{\mu }_{0}}i}{\pi L}\]                                    

D)  \[\frac{\sqrt{2}{{\mu }_{0}}i}{\pi L}\]

Correct Answer: C

Solution :

The magnetic field at 0 use to wires AB and DE is zero and that due to wires BC and CD is\[\frac{{{\mu }_{0}}i}{4\pi (L/2)}\]. \[(\sin {{45}^{o}}+\sin {{45}^{o}})=\frac{{{\mu }_{0}}i}{2\pi L}\]each The net magnetic field\[=2\times \frac{{{\mu }_{0}}i}{2\pi L}=\frac{{{\mu }_{0}}i}{\pi L}\].