Question

The ratio of the magnetic field at the centre of a circular loop to the magnetic field at the centre of the square loop, which is made by a constant length current carrying wire is:

• A. $\frac{{\pi }^2}{16}$
• B. $\frac{{\pi }^2}{8\sqrt{2}}$
• C. $\frac{{\pi }^2}{4\sqrt{2}}$
• D. $\frac{{\pi }^2}{2\sqrt{2}}$

Answer 1

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

circular loop

Length $L=2\pi r$\

Magnetic field $B=\frac{{\mu }_0}{4\pi }\frac{2\pi i}{r}=\frac{{\mu }_0}{4\pi }\frac{4{\pi }^{2}i}{r}$

square loop

Length $L=4a$

$B=4\times \frac{{\mu }_0}{4\pi }\frac{2\sqrt{2i}}{a}$

$B=\frac{{\mu }_0}{4\pi }\frac{8\sqrt{2i}}{\left(\frac{1}{2}\pi r\right)}$

$\frac{B_{circular}}{B_{square}}=\frac{\frac{{\mu }_0}{4\pi }\frac{4{\pi }^{2}i}{r}}{\frac{{\mu }_0}{4\pi }\frac{8\sqrt{2i}}{\left(\frac{1}{2}\pi r\right)}}=\frac{{\pi }^2}{8\sqrt{2}}$