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 will be.

• A. $\frac{{\pi }^2}{4\sqrt{2}}$
• B. $\frac{{\pi }^2}{8\sqrt{2}}$
• C. $\frac{{\pi }^2}{6\sqrt{2}}$
• D. $\frac{{\pi }^2}{2\sqrt{2}}$

Answer 1

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

For a circular loop,

The magnetic field at the center of the circular coil is

$B_{cir}=\frac{{\mu }_{0}i}{2r}$. . . . . .(1)

Now, $2\pi r=4a$

$a=\frac{\pi r}{2}\leftarrow$ side of the square

For square loop, the magnetic field at the center of the square due to 4 side is

$B_{squ}=4\times \frac{{\mu }_{0}i}{4\pi a/2}(sin{45}^0+sin{45}^0)$

$B_{squ}=\frac{4{\mu }_{0}i}{4\pi .\frac{{\pi }^2{r}^2}{4}}\left(\sqrt{2}\right)=\frac{4\sqrt{2}}{{\pi }^2}\frac{{\mu }_{0}i}{r}$. . . . . . ..(2)

Hence, $\frac{B_{cir}}{B_{squ}}=\frac{\frac{{\mu }_{0}i}{2r}}{\frac{4\sqrt{2}}{{\pi }^2}\frac{{\mu }_{0}i}{r}}$

$\frac{B_{cir}}{B_{squ}}=\frac{{\pi }^2}{8\sqrt{2}}$

The correct option is B.