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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 (4 μ o ia 2 )/ π (4x 2 +a 2 ) (4x 2 +2a 2 ) 1/2 π is pi.

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)1/2


π is pi.



 

Grade:upto college level

1 Answers

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,

askIITians Faculty

B.Tech.

IIT Kanpur

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