1. Displacement Current and the Generalized Ampere’s Law AP Physics C Montwood High School R. Casao

2. Charges in motion, or currents, produce magnetic fields. When a current-carrying conductor has high symmetry, we can determine the magnetic field using Ampere’s law: – where the line integral is over any closed path through which the conduction current passes. –Conduction current is the current carried by the wire. –The conduction current is defined by:

3. Ampere’s law in this form is only valid if the conduction current is constant over time. This means that any electric fields present are constant as well. Maxwell recognized this limitation of Ampere’s law and modified the law to include electric fields that change over time. Consider a charging capacitor: –The current I is decreasing over time as the magnitude of the charge on the capacitor and the electric field between the capacitor plates increases.

4. –No conduction current passes between the capacitor plates. Consider the two Amperian surfaces S 1 and S 2 bounded by the same path P. –Ampere’s law says that the line integral of Bds around this path must equal µ o ·I, where I is the total current through any surface bounded by the path P.

5. When the path P bounds S 1, the result of the integral is µ o ·I since the conduction current passes through S 1. When the path bounds S 2, the result of the integral is zero since no conduction current passes through S 2. Current passes through S 1 but does not pass through S 2.

6. Maxwell solved this problem by adding a displacement current I d to the right side of Ampere’s law: Displacement current I d is proportional to the rate of change of the electric flux  E. Electric flux  E is defined as:

7. As the capacitor is being charged (or discharged), the changing electric field between the plates can be considered as a current that bridges the discontinuity in the conduction current. Adding the displacement current to the right side of Ampere’s law allows some combination of conduction current and displacement current to pass through surfaces S 1 and S 2. Ampere-Maxwell law:

8. The electric flux through S 2 is: –E is the uniform electric field between the plates. –A is the area of the plates. The conduction current passes through S 1.

9. If the charge on each plate at any instant is Q, then: The electric flux through S2 is:

10. The displacement current I d through S 2 is: The conduction current is equal to the displacement current: I = I d Magnetic fields are produced by both conduction currents and changing electric fields.

11. Displacement Current in a Capacitor An alternating current (AC) voltage is applied directly across an 8  F capacitor. The frequency of the AC source is 3 kHz and the voltage amplitude (V max ) is 30 V. Determine the displacement current between the plates of the capacitor. Angular frequency of the oscillation is:  = 2·  ·f Voltage as a function of time: V = V max ·sin(  ·t)

12. Charge on the capacitor is Q = C·V

13. Substituting: The displacement current graphs as a sine wave with a maximum value of 4.5239 A.