Question

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$\mathit{P}\mathbf{\text{'}}$.

Short answer

(a) Direction of field at a point$P$ is along negative x axis , Direction of field at a point $P\text{'}$is along positive x axis

(b) We can show that$B=\frac{1}{2}·{\mu }_0\lambda$

Step-by-step solution

Given

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

Understanding the concept

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:

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

(a) Calculate the direction of field at a point P and Direction of field at a point   P'

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.

(b) Show that B=12·μ0λ

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.