PHYS208 Fundamentals of Physics II

Hint for Ch. 30, 45P

For the amperian loop indicated in the drawing, ienc is zero. There will be a contribution to the line integral of Ampere's law from the leftmost straight section but from none of the other sections. How can this be?

Use a circular amperian loop of radius r centered about the axis of the coax cable. The magnetic field found from application of Ampere's law will be (mu0 ienc)/(2 pi r). The field in each region will then depend on the current enclosed by the amperian loop with that radius.

Hint for Ch. 30, 48P

Application of Ampere's law proceeds as in Problem 47 part a) above. However the current enclosed by the amperian loop must be calculated by integrating the current density over the area enclosed.

Hint for Ch. 30, 49P

The field outside the circular pipe will behave as if its current is concentrated on its axis.

Recognize that the unknown current in the wire must point in the same direction as the current in the pipe to satisfy the stated critera.

Hint 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. What happens to the field?

Now rotate the sheet by 180 degrees about a line perpendicular to the sheet. What happens to the field? Can you see a contradiction?

b) Use a rectangular loop oriented similar to that used for solution of the solenoid configuration.


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Last updated April 2, 1998.
Copyright George Watson, Univ. of Delaware, 1997.