Question

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 is

• A. $\frac{\pi }{2\sqrt{2}}$
• B. $\frac{\pi }{4\sqrt{2}}$
• C. $\frac{{\pi }^2}{4\sqrt{2}}$
• D. $\frac{{\pi }^2}{8\sqrt{2}}$

Answer 1

The correct option is D $\frac{{\pi }^2}{8\sqrt{2}}$

Magnetic field at the centre of a circular coil
$B_{Circular}=\frac{{\mu }_0}{4\pi }\frac{2\pi i}{r}=\frac{{\mu }_0}{4\pi }\frac{4{\pi }^{2}i}{L}[\because L=2\pi randr=\frac{1}{2\pi }]$
Magnetic field at the centre of square coil
$B_{Square}=\frac{{\mu }_0}{4\pi }\frac{8\sqrt{2}i}{a}=\frac{{\mu }_0}{4\pi }\frac{32\sqrt{2}i}{L}[\because L=4aanda=\frac{L}{4}]$
Hence, $\frac{B_{Circular}}{B_{Square}}=\frac{\frac{m{u}_0}{4\pi }\frac{4{\pi }^{2}i}{L}}{\frac{{\mu }_0}{4\pi }\frac{32\sqrt{2}i}{L}}=\frac{{\pi }^2}{8\sqrt{2}}$.