question_answer

The wire loop formed by joining two semicircular sections of radii R1 and R2 and centre C, carries a current I as shown. The magnetic field at C has a magnitude:

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

B)  \[\frac{{{\mu }_{0}}I}{4}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]

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

D)  \[\frac{{{\mu }_{0}}I}{4}\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\]

Correct Answer: B

Solution :

 Magnetic field at C due to AB and DE = zero Magnetic field due to semicircular arc \[BPD=\frac{{{\mu }_{0}}I}{4\pi {{R}_{2}}}\](downward) Magnetic field due to semicircular arc (upward)        \[{{B}_{net}}=\left( \frac{{{\mu }_{0}}I}{4{{R}_{1}}}-\frac{{{\mu }_{0}}I}{4{{R}_{2}}} \right)\](upward)