Q. The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be
A) $\frac{\pi^{2}}{4\sqrt{2}}$
B) $\frac{\pi^{2}}{8\sqrt{2}}$
C) $\frac{\pi}{2\sqrt{2}}$
D) $\frac{\pi}{4\sqrt{2}}$
Explanation
(b) Magnetic field at the centre of circular coil
$B_{circular} = \frac{\mu_{0}}{4\pi}.\frac{2\pi i}{r} = \frac{\mu_{0}}{4\pi}.\frac{4\pi^{2}i}{L}$
Magnetic field at the centre of square coil
$B_{square} = \frac{\mu_{0}}{4\pi}.\frac{8\sqrt{2}i}{a} = \frac{\mu_{0}}{4\pi}.\frac{32\sqrt{2}i}{L}$
Hence $\frac{B_{circular}}{B_{square}} = \frac{\pi^{2}}{8\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.