# Show that the magnetic field at a point on the axis of a square loop( radius: a; at a distance  x from the centre of the loop) is (4μoia2)/ π(4x2+a2) (4x2+2a2)1/2π is pi.

Saurabh Singh
askIITians Faculty 49 Points
10 years ago

Start with the Biot-Savart law for linear currents:

$\displaystyle \mathbf{B}\left(\mathbf{r}\right)=\frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{l}^{\prime}\times\left(\mathbf{r}-\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|^{3}}$

Now consider the edge that extends from${y=-w/2}$to${y=+w/2}$at${x=+w/2}$. Along this edge we have

 $\displaystyle \mathbf{r}-\mathbf{r}'$ $\displaystyle =$ $\displaystyle -x^{\prime}\hat{\mathbf{x}}-y^{\prime}\hat{\mathbf{y}}+z\hat{\mathbf{z}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{w}{2}\hat{\mathbf{x}}-y^{\prime}\hat{\mathbf{y}}+z\hat{\mathbf{z}}$ $\displaystyle d\mathbf{l}^{\prime}$ $\displaystyle =$ $\displaystyle dy^{\prime}\hat{\mathbf{y}}$ $\displaystyle d\mathbf{l}^{\prime}\times\left(\mathbf{r}-\mathbf{r}'\right)$ $\displaystyle =$ $\displaystyle \left(z\hat{\mathbf{x}}+\frac{w}{2}\hat{\mathbf{z}}\right)dy^{\prime}$

The integral is then

 $\displaystyle \mathbf{B}_{1}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{4\pi}\left(z\hat{\mathbf{x}}+\frac{w}{2}\hat{\mathbf{z}}\right)\int_{-w/2}^{w/2}\frac{dy^{\prime}}{\left(\frac{w^{2}}{4}+y^{\prime2}+z^{2}\right)^{3/2}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{4\pi}\left(z\hat{\mathbf{x}}+\frac{w}{2}\hat{\mathbf{z}}\right)\frac{8w}{\sqrt{2w^{2}+4z^{2}}\left(w^{2}+4z^{2}\right)}$

By symmetry, the opposite edge at${x=-w/2}$will contribute the same${z}$component and the opposite${x}$component. Similarly, the two edges at${y=\pm w/2}$will contribute two more${z}$components with the${y}$components cancelling. Thus the total field is

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle 4\frac{\mu_{0}I}{4\pi}\frac{w}{2}\frac{8w}{\sqrt{2w^{2}+4z^{2}}\left(w^{2}+4z^{2}\right)}\hat{\mathbf{z}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4w^{2}\mu_{0}I}{\pi\sqrt{2w^{2}+4z^{2}}\left(w^{2}+4z^{2}\right)}\hat{\mathbf{z}}$

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.37.

Thanks & Regards

Saurabh Singh,

B.Tech.

IIT Kanpur