Read through the course notes before watching the video. The course note files may also contain links to associated animations or interactive simulations.
Read section 13.1:
Maxwell’s Equations and Electromagnetic Waves (PDF - 1.1MB)
Do the Concept Questions first to make sure you understand the main concepts from this module. Then, when you are ready, try the Challenge Problems.
Watch the Problem Solving Help videos for insights on how to approach and solve problems related to the concepts in this module.
What does the displacement current term added to Ampere’s Law mean in practice? Discuss this in terms of an open surface and the contour bounding that surface. Apply this meaning to the classic problem of a charging capacitor.
Download this video:
» iTunes U (MP4 - 59MB)
» Internet Archive (MP4 - 59MB)
A wire with a circular cross-sectional area of 40 square millimeters carries a current of 30 A. The resistivity of the wire is 2 X 10-8 Ohm-meter. What is the uniform electric field in the wire? If the current changes at a rate of 6000 Amps/second, what rate is the electric field changing? What is the displacement current density in the material? What is the magnetic field 5 centimeters from the center of the wire? Note that you must include both the displacement current and the conduction current in this calculation. Is the contribution from the displacement current in the case important?
Download this video:
» iTunes U (MP4 - 22MB)
» Internet Archive (MP4 - 22MB)
Consider a spherical Gaussian surface with a little hole cut out of the top, so that the surface is bounded by a small Amperean loop shaped like a tiny latitude circle on a globe. Suppose that the sphere contains a net charge Qencl, which may be changing in time owing to current flowing into or out of the sphere. Show that, in the limit that the hole is shrunk to zero size, Maxwell’s equations imply that
dQencl/dt + Ithru = 0
where Ithru is the net current going outward anywhere through the Gaussian surface. Explain what this equations means in 25 words or less. What would happen to this conclusion if the displacement current were not included in Ampere’s Law? Why is this nonsense? Maxwell actually turned this argument around and used it to justify the displacement current term.
Download this video:
» iTunes U (MP4 - 13MB)
» Internet Archive (MP4 - 13MB)
The visualizations linked below are related to the concepts covered in this module.