Question

A wide, long, insulating belt has a uniform positive charge per unit area s on its upper surface. Rollers at each end move the belt to the right at a constant speed v. Calculate the magnitude and direction of the magnetic field produced by the moving belt at a point just above its surface. (Hint:At points near the surface and far from its edges or ends, the moving belt can be considered to be an infinite current sheet like that in Problem 28.73.)

Short answer

The magnitude of the magnetic field is $\frac{P_{0}v\sigma }{2}$ and the direction is outside the page.

Step-by-step solution

 Identification of the concept

The conveyer belt is considered or approximated in this case to be an infinitely long current sheet.

Also from the mentioned problem 28.73,the magnetic field B of an infinite current sheet is,

$B=\frac{1}{2}{\mu }_0\ln \mid$

Here, symbols have their usual meanings and $n=\frac{1}{L}$, L being the width of the sheet.

Determination of the magnitude and direction of the magnetic field produced by the conveyer belt.

Consider a small length on the conveyer belt to be $∆x$.

Therefore charge in that small length element is,

$\Delta Q=L\Delta x\sigma$

Now, the current expression is,

$\begin{array}{r}\mathrm{I}=\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}\\ =L\frac{\mathrm{\Delta x}}{\mathrm{\Delta t}}\sigma \\ =L\mathrm{V}\sigma \end{array}$

Thus, by approximating the same magnetic field as an infinite sheet,

$\begin{array}{r}B=\frac{{\mu }_0}{2\mathrm{L}}\mathrm{I}\\ =\frac{{\mu }_0\mathrm{v}\sigma }{2}\end{array}$

Thus the magnitude is obtained. Refer to the image below for checking the direction of the magnetic field.

The direction of the magnetic field is out of the page.