question_answer

In the loop shown, the magnetic induction at the point O is    

A)  \[\frac{{{\mu }_{0}}I}{8}\left( \frac{{{R}_{1}}-{{R}_{2}}}{{{R}_{1}}{{R}_{2}}} \right)\]

B)   \[\frac{{{\mu }_{0}}I}{8}\left( \frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}{{R}_{2}}} \right)\]

C)  \[\frac{{{\mu }_{0}}I}{8}\left( \frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\]

D)  zero

Correct Answer: B

Solution :

:       Since the point 0 lies on the lines of the straight wires, they contribute no magnetic induction at 0. The magnetic induction at 0 due to arcs only. The magnetic induction at 0 due to arc of radius \[{{R}_{1}}\] is \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{{{R}_{1}}}\frac{\pi }{2}=\frac{{{\mu }_{0}}I}{8{{R}_{1}}}\otimes \] The magnetic field at O due to arc of radius \[{{R}_{2}}\] is \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{{{R}_{2}}}\frac{\pi }{2}=\frac{{{\mu }_{0}}I}{8{{R}_{2}}}\otimes \] The net magnetic field at O is \[B={{B}_{1}}+{{B}_{2}}\] \[B=\frac{{{\mu }_{0}}I}{8{{R}_{1}}}+\frac{{{\mu }_{0}}I}{8{{R}_{2}}}\] \[=\frac{{{\mu }_{0}}I}{8}\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)=\frac{{{\mu }_{0}}I}{8}\left( \frac{{{R}_{2}}+{{R}_{1}}}{{{R}_{1}}{{R}_{2}}} \right)\]