Question
The magnetic field B produced by the current i flowing through the sides of a square loop of side ℓ, at a point P at distance x from the center of the square, on the axis perpendicular to the plane of the square loop and passing through its center, is
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Solution
[Key Concept: Field of a finite straight wire, B = (μ0i/4πd)(sin α1 + sin α2); for the loop only the axial components of the four sides add.]
B due to AB
$$\begin{matrix} & dB = \frac{\mu_{0}I}{4\pi}\frac{\overrightarrow{\text{ }d\mathcal{l}} \times \overrightarrow{r}}{r^{3}} \\ & \overrightarrow{\text{ }d\mathcal{l}} = dz\widehat{k} \end{matrix}$$
Position of element $\left( 0,\frac{\mathcal{l}}{2},z \right)$
$$\begin{matrix} & \overrightarrow{r} = {\overrightarrow{r}}_{p} - {\overrightarrow{r}}_{\text{element~}} \\ & \overrightarrow{r} = x\widehat{i} - \left( \frac{\mathcal{l}}{2}\widehat{j} + z\widehat{k} \right) \\ & \overrightarrow{r} = x\widehat{i} - \frac{\mathcal{l}}{2}\widehat{j} - z\widehat{k} \\ & \ |\overrightarrow{r}| = \sqrt{x^{2} + \frac{\mathcal{l}^{2}}{4} + z^{2}} \\ & d\overrightarrow{B} = \frac{\mu_{0}I}{4\pi}\frac{dz\widehat{k} \times \left( x\widehat{i} - \frac{\mathcal{l}}{2}\widehat{j} - z\widehat{k} \right)}{\left( x^{2} + \frac{\mathcal{l}^{2}}{4} + z^{2} \right)^{\frac{3}{4}}} \\ & d\overrightarrow{B} = \frac{\mu_{0}I}{4\pi}\frac{\left\lbrack xdz\widehat{j} + \frac{\mathcal{l}}{2}dz\widehat{i} \right\rbrack}{\left( x^{2} + \frac{\mathcal{l}^{2}}{4} + z^{2} \right)^{\frac{3}{4}}} \end{matrix}$$
Taking x-component
$$\begin{matrix} & \ |dB|_{x} = \frac{\mu_{0}I}{4\pi}\frac{\mathcal{l}}{2}\frac{dz}{\left( x^{2} + \frac{\mathcal{l}^{2}}{4} + z^{2} \right)^{\frac{3}{4}}} \\ & {\text{ }B}_{x} = \frac{\mu_{0}I\mathcal{l}}{8\pi}\int_{- \frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}\mspace{2mu}\mspace{2mu}\frac{dz}{\left( x^{2} + \frac{\mathcal{l}^{2}}{4} + z^{2} \right)^{\frac{3}{4}}} \end{matrix}$$
Consider $x^{2} + \frac{\mathcal{l}^{2}}{4} = r^{2}$
$$B_{x} = \frac{\mu_{0}I\mathcal{l}}{8\pi}\int_{- \frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}\mspace{2mu}\mspace{2mu}\frac{dz}{\left( r^{2} + z^{2} \right)^{\frac{3}{4}}}$$
$$\begin{matrix} & \ \\ B_{x} & \ = \frac{\mu_{0}I\mathcal{l}^{2}}{8\pi\left( x^{2} + \frac{\mathcal{l}^{2}}{4} \right)\sqrt{x^{2} + \frac{\mathcal{l}^{2}}{2}}} \end{matrix}$$
Net $B = 4{\text{ }B}_{x} = \frac{4 \times 4\mu_{0}I\mathcal{l}^{2} \times \sqrt{2}}{8\pi\left( 4x^{2} + \mathcal{l}^{2} \right)\sqrt{2x^{2} + \mathcal{l}^{2}}}$
$$B = \frac{8\sqrt{2}\mu_{0}I\mathcal{l}^{2}}{4\pi\left( 4x^{2} + \mathcal{l}^{2} \right)\sqrt{2x^{2} + \mathcal{l}^{2}}}$$
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