Question

Two wires, both of length $\mathit{L}$, are formed into a circle and a square, and each carries current$\mathit{i}$. Show that the square produces a greater magnetic field at its center than the circle produces at its center.

Short answer

From the calculation, we can show that the square produces a greater magnetic field at the centre than the circle.

Step-by-step solution

Given Data

Two wires are of length, and the current in both the wires is same.

Understanding the concept

We use the formula of magnetic field due to an infinitely long wire at perpendicular distance to calculate the magnetic field due to the wires. Also, we use the formula of magnetic field at the centre of circle, and comparing both results, we conclude that the magnetic field at the centre of the square is greater than that of the circle.

Formulae:

For square,

${\mathit{B}}_{\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{r}\mathbf{e}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{4}\mathbf{\pi }\mathbf{R}}\left(sin{\theta }_1+sin{\theta }_2\right)$

For circle,

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{R}}$

Show that the square produces a greater magnetic field at the center than the circle.

Let the length of each side of the square be $a$, and the center of the square be at a distance from any side be $\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

The magnitude of magnetic field at the center due to any one of the wires is given by

$B_{centre}=\frac{{\mu }_{0}i}{4\pi R}\left(sin{\theta }_1+sin{\theta }_2\right)$

Where ${\theta }_1,{\theta }_2$are the angles made at the center of the square, and $R$is the distance of the wire from the center of the square.

$R=\frac{a}{2}$and angles ${\theta }_1,{\theta }_2=\theta$

$B_{centre}=\frac{{\mu }_{0}i}{4\pi R}\left(2sin\theta \right)$

$$\begin{gathered} B_{centre}=\frac{{\mu }_{0}i}{4\pi \left(\frac{a}{2}\right)}\frac{2\left(\frac{a}{2}\right)}{\frac{\sqrt{2}a}{2}} \\ =\frac{{\mu }_{0}i}{2\pi a}\frac{a}{\frac{\sqrt{2}a}{2}} \\ =\frac{{\mu }_{0}i}{\sqrt{2}\pi a} \end{gathered}$$

This is the magnetic field for one wire. Therefore, for 4 wires, multiply the above result by 4.

$$\begin{gathered} B_{centre}=\frac{4{\mu }_{0}i}{\sqrt{2}\pi a} \\ =\frac{2\sqrt{2}{\mu }_{0}i}{\pi a} \end{gathered}$$ ...(i)

The magnetic field at the center of the circular wire of radius (eq.29-10) is given by

$B=\frac{{\mu }_{0}i}{2R}$ ...(ii)

From equations (i) and (ii) we can say that

$\frac{2\sqrt{2}{\mu }_{0}i}{\pi a}>\frac{{\mu }_{0}i}{2R}$

$\frac{4\sqrt{2}}{\pi a}>\frac{1}{R}$ ...(iii)

For a square, the total length is $4a$, and for a circle, the total length is $2\mathrm{\pi R}$. We know that both wires are of the same length.

Hence,$4a=2\pi R$

$a=\frac{\pi R}{2}$

Substituting this value in equation (iii), we get,

localid="1662996602692" $\frac{4\sqrt{2}}{\pi a}=\frac{8\sqrt{2}}{{\pi }^{2}R}>\frac{1}{R}$

From the above result, we can say that the square produces a greater magnetic field at the center than the circle.