Q. A current $I$ enters a circular coil of radius $R,$ branches into two parts and then recombines as shown in the circuit diagram The resultant magnetic field at the centre of the coil is
A) Zero
B) $\frac{\mu_{0}I}{2R}$
C) $\frac{3}{4}\left( \frac{\mu_{0}I}{2R} \right)$
D) $\frac{1}{4}\ \left( \frac{\mu_{0}I}{2R} \right)$
Explanation
(a) Magnetic field a $B$
$B_{1} = \frac{\mu_{0}}{2\pi}\frac{i\mathrm{\Delta}l}{r}$
Magnetic field at $A$
$B_{2} = \frac{\mu_{0}}{2\pi}\frac{i\mathrm{\Delta}l}{r}$
The resultant magnetic field at the centre
$\left| B_{A} \right| = \left| B_{B} \right|$
So, magnetic field become is zero.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.