Question

An infinitely wide flat sheet of charge flows out of the page in FIGURE CP29.83. The current per unit width along the sheet (amps per meter) is given by the linear current density Js. a. What is the shape of the magnetic field? To answer this question, you may find it helpful to approximate the current sheet as many parallel, closely spaced current-carrying wires. Give your answer as a picture showing magnetic field vectors. b. Find the magnetic field strength at distance d above or below the current sheet.

Short answer

Part a

The shape of the magnetic field is horizontal and to the left above the sheet; horizontal and to the right below the sheet.

Part b

The magnetic field strength at a distance $d$above or below the current sheet is$\frac{1}{2}{\mu }_0{J}_s$.

Step-by-step solution

Given information

An infinitely wide flat sheet of charge flows out of the page is shown below

Part a

The current sheet extends to infinity in both directions.

The shape of the magnetic field is consistent with the symmetry of the current sheet.

Consider two current-carrying wires.

The magnetic field above the midpoint of the wires is horizontal and to the left.

The magnetic field below the midpoint of the wires is horizontal and to the right.

If the current sheet consists of many such pairs of wires, closely spaced, then the magnetic field above the sheet is everywhere horizontal and to the left while the magnetic field below the sheet is everywhere horizontal and to the right.

The above description is shown in the figure below

Therefore, the shape of the magnetic field is horizontal and to the left above the sheet; horizontal and to the right below the sheet.

Part b

The closed path of width $L$is shown in the figure is parallel to the field along the top and bottom edges, perpendicular to the field along the left and right edges.

Thus the line integral in Ampere’s law is

$$\begin{gathered} \oint \overrightarrow{B}·\overrightarrow{ds}={\int }_{top}\overrightarrow{B}·\overrightarrow{ds}+{\int }_{left}\overrightarrow{B}·\overrightarrow{ds}+{\int }_{bottom}\overrightarrow{B}·\overrightarrow{ds}+{\int }_{right}\overrightarrow{B}·\overrightarrow{ds} \\ =BL+0+BL+0 \\ =2BL \end{gathered}$$

As the current per unit width is $J_s$, the amount of current in the length $L$of the current sheet is $J_{s}L$.

By Ampere's law,

$$\begin{gathered} \oint \overrightarrow{B}·\overrightarrow{ds}={\mu }_{0}I \\ 2BL={\mu }_0{J}_{s}L \\ B=\frac{1}{2}{\mu }_0{J}_s \end{gathered}$$

Therefore, the magnetic field strength at a distance $d$above or below the current sheet is $\frac{1}{2}{\mu }_0{J}_s$.