A straight conductor carrying a current I splits into two identical semicircular arcs as shown in figure below. What is the magnitude of the magnetic field at the centre C of the resulting circular arcs of radius R?
A) 0
B) \[\frac{{{\mu }_{0}}l}{4R}\]
C) \[\frac{{{\mu }_{0}}l}{2R}\]
D) \[\frac{{{\mu }_{0}}l}{R}\]
Correct Answer:
A
Solution :
Magnetic field due to BAC arc is \[B=\frac{{{\mu }_{0}}\pi (I/2)}{4\pi R}\otimes \] ... (i) Magnetic field due to ADC arc is \[B=\frac{{{\mu }_{0}}\pi (I/2)}{4\pi R}\] ? (ii) From Eqs. (i) and (ii) net magnetic field is zero.