Magnetism> square-loop...

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 (4μ_oia²)/ π(4x²+a²) (4x²+2a²)^1/2

π is pi.

Pranjal K , 11 Years ago

Grade upto college level

1 Answers

Saurabh Singh

Last Activity: 11 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

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