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 of one turn made from the same length of wire will be

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

Answer 1

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

Let the length of the wire be $L$

Case -1 : Circular coil

$L=2\pi r\Rightarrow r=\frac{L}{2\pi }$

Magnetic field, $B=\frac{{\mu }_{0}i}{2r}=\frac{\pi {\mu }_{0}i}{L}$

$B_C=\frac{\pi {\mu }_{0}i}{L}$

Case - 2 : Square coil


$4a=L$

$\Rightarrow a=\frac{L}{4}$

Magnetic field due to each section/side of square shaped coil,

$B=\frac{{\mu }_{0}i}{4\pi d}[\sin {45}^{\circ }+sin{45}^{\circ }]$

or, $B=\frac{{\mu }_{0}i}{4\pi \left(\frac{a}{2}\right)}[2\times \frac{1}{\sqrt{2}}]$

$B=\frac{2{\mu }_{0}i}{4\pi a}\times \frac{2}{\sqrt{2}}$

or, $B=\frac{4{\mu }_{0}i}{\sqrt{2}\pi L}$

Thus net magnetic field at centre of square is,

$B_S=4\times \frac{4{\mu }_{0}i}{\sqrt{2}\pi a}=\frac{16{\mu }_{0}i}{\sqrt{2}\pi L}$

Now, $\frac{B_c}{B_s}=\frac{\left(\frac{\pi {\mu }_{0}i}{L}\right)}{\left(\frac{16{\mu }_{0}i}{\sqrt{2}\pi L}\right)}$

$\Rightarrow \frac{B_c}{B_s}=\frac{\sqrt{2}{\pi }^2}{16}=\frac{2{\pi }^2}{16\sqrt{2}}$

$\therefore \frac{B_c}{B_s}=\frac{{\pi }^2}{8\sqrt{2}}$