question_answer

A wire carrying current\[i\]is shaped as shown. Section AB is a quarter circle of radius r. The magnetic field is directed:

A)  perpendicular to the plane of the paper and directed into the paper

B)  at an angle \[\pi /4\]to the plane of the paper

C)  along the bisector of the angle ACB away from AB

D)  along the bisector of ACB towards AB.

Correct Answer: A

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