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Displacement Current and the Generalized Ampere’s Law AP Physics C Montwood High School R. Casao • 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: B ds μ I o enclosed – where the line integral is over any closed path through which the conduction current passes. – Conduction current is the current carried by the wire. dQ – The conduction current is defined by: I dt • 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. – No conduction current passes between the capacitor plates. • Consider the two Amperian surfaces S1 and S2 bounded by the same path P. – Ampere’s law says that the line integral of B•ds around this path must equal µo·I, where I is the total current through any surface bounded by the path P. • When the path P bounds S1, the result of the integral is µo·I since the conduction current passes through S1. • When the path bounds S2, the result of the integral is zero since no conduction current passes through S2. • Current passes through S1 but does not pass through S2. • Maxwell solved this problem by adding a displacement current Id to the right side of Ampere’s law: B ds μ I o enclosed Id • Displacement current Id is proportional to the rate of change of the electric flux E. dΦ E Id ε o dt • Electric flux E is defined as: Φ E E dA • 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 S1 and S2. • Ampere-Maxwell law: B d s μ o I conduction I displacement dΦ E B d s μ o I conduction μ o ε o dt • The electric flux through S2 is: Φ E E dA E A – E is the uniform electric field between the plates. – A is the area of the plates. • The conduction current passes through S1. • If the charge on each plate at any instant is Q, then: Q εo A V C C E V d d εo A Q Q E d Ed εo A • The electric flux through S2 is: Q Q A ΦE E A εo εo A • The displacement current Id through S2 is: Q d εo dΦ E Id ε o εo dt dt 1 dQ dQ Id ε o ε o dt dt • The conduction current is equal to the displacement current: I = Id • Magnetic fields are produced by both conduction currents and changing electric fields. 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 (Vmax) 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 = Vmax·sin( ·t) • Charge on the capacitor is Q = C·V dQ d(C V) dV Id C dt dt dt dV dVmax sin ω t dsin ω t Vmax dt dt dt dV Vmax ω cosω t dt I d C Vmax ω cosω t • Substituting: ω 2 π f 2 π 3000Hz 6000 π Hz I d C Vmax ω cos6000 π Hz t I d 8 x 106 F 30 V 6000 π Hz cos6000 π Hz t I d 4.5239 A cos6000 π Hz t • The displacement current graphs as a sine wave with a maximum value of 4.5239 A.