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)You need to login to perform this action.
You will be redirected in
3 sec