Figure 29-84 shows a cross section of an infinite conducting sheet carrying a current per unit x-length of $\mathit{\lambda }$; the current emerges perpendicularly out of the page. (a) Use the Biot – Savart law and symmetry to show that for all pointsP above the sheet and all points $\mathit{P}\mathbf{\text{'}}$below it, the magnetic field$\overrightarrow{\mathbf{B}}$is parallel to the sheet and directed as shown. (b) Use Ampere’s law to prove that $\mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{·}{\mathit{\mu }}_{\mathbf{0}}\mathit{\lambda }$ at all points P and$\mathit{P}\mathbf{\text{'}}$.

• Current per unit length is$\lambda =I/L_x$
• I is the current through sheet

Using the right hand thumb rule, we can find the direction of the field produced by the current carrying wires. Using Ampere’s law, we can find the field in terms of current per unit length.

Formula:

$\oint \overrightarrow{B}·\overrightarrow{ds}={\mu }_0{I}_{enclosed}$

Referring fig 29-84, from the direction of current and using right hand thumb rule, we get thedirection of the field at point$P$along negativex axis. Direction of Field at a point$P\text{'}$is along positive x axis.

Now to show the second part, consider Amperes law, let’s assume that width of the wire along x direction is$L_x$then

$\begin{array}{rcl}\oint \overrightarrow{B}·\overrightarrow{ds}& =& {\mu }_0{I}_{enclosed}\\ B\oint dx& =& {\mu }_0\lambda L_x\\ B·L_x& =& {\mu }_0\lambda L_x\\ B& =& {\mu }_0\lambda \end{array}$

By considering the upper and lower part of sheet, and by symmetry, we have half contribution of original so that$B=\frac{{\mu }_0\lambda }{2}$

Hence it is proved.