PHYS208 Fundamentals of Physics II

Answer for Ch. 30, 51P

a) Suppose the field is not parallel to the sheet, as shown:

[current sheet; B arbitrary]

Reverse the direction of the current. By the Biot-Savart law, B should reverse direction.

[current sheet reversed]

Now rotate the sheet by 180 degrees about a line perpendicular to the sheet. B should now point down rather than up, for the same current configuratino as first shown -- in contradication to our initial assumption.

[current sheet restored]

The only direction of B which is preserved under these transformations is one directed horizontally.

b) Evaluate the line integral of Ampere's law around the rectangular loop shown. By the symmetry of the current sheet, B will have the same magnitude at the same distance above and below the sheet; the upper and lower sides of the amperian loop are constructed to be the same distance from the sheet, each having length a.

[current sheet - Ampere's law]

If the line integral is evaluated counterclockwise, the contribution of each of these sides is Ba. There will be no contribution from the other two sides, since B is perpendicular at all points. So the line integral around the amperian loop is 2Ba.

The enclosed current is the linear current density lambda times the length a of the side of the loop; ienc = lambda a.

Thus by Ampere's law, B = mu0 lambda/2, directed as shown.

Last updated Oct. 30, 1997.
Copyright George Watson, Univ. of Delaware, 1997.