ELEC 3105 Basic EM and Power Engineering Faraday’s Law Lenz’s Law Displacement Current
Introduction: So far we have Faraday’s Law Introduction: So far we have These equations are OK for static fields, i.e. those fields independent of time. When fields vary as a function of time the curl equations acquire an additional term. gets a gets a
ELEC 3105 Basic EM and Power Engineering Faraday’s Law
Faraday’s Law Consider the following experiment. Pull a wire loop through a region of non-uniform magnetic field . Wire loop, path encloses a surface S area gained in time area lost in time Magnetic field vector points into the page
Faraday’s Law Consider the following experiment. Pull a wire loop through a region of non-uniform magnetic field . Charge in wire Charge in wire Magnetic field vector points into the page
Faraday’s Law 𝑊 𝑞 =𝑉 Early definition of potential Consider the work done on +1C test charge moved around the loop; this is the “emf” electromotive force. Examine Expression Charge in wire Charge in wire
Faraday’s Law Consider the work done on +1C test charge moved around the loop; this is the “emf” electromotive force. Now: Flux change at right side of loop Flux change at left side of loop Note that B and da are in opposite directions
Faraday’s Law This is a general result. Even if we hold the loop stationary and change B, the emf is still given by the negative rate of change of the flux. N S Move magnet
The induced emf always opposes the change in Faraday’s Law Further generalization is possible. Consider moving loop in time varying magnetic field. The induced emf always opposes the change in flux MOTIONAL emf TRANSFORMER emf N S Move magnet Move loop
ELEC 3105 Basic EM and Power Engineering Lenz’s Law
The induced emf always opposes the change in flux Faraday’s Law / Lenz’s Law The induced emf always opposes the change in flux motional emf Move loop towards magnet B increases in loop Flux increases in loop Current induced through emf Current produces magnetic field in loop: 2nd postulate This magnetic field in opposite in direction to magnetic field of magnet I N S MOVE LOOP
The induced emf always opposes the change in flux Faraday’s Law / Lenz’s Law The induced emf always opposes the change in flux motional emf Move loop away from magnet B decreases in loop Flux decreases in loop Current induced Current produces magnetic field in loop This magnetic field in same direction to magnetic field of magnet I N S MOVE LOOP
The induced emf always opposes the change in flux Faraday’s Law / Lenz’s Law The induced emf always opposes the change in flux transformer emf Move magnet towards loop B increases in loop Flux increases in loop Current induced Current produces magnetic field in loop This magnetic field in opposite direction to magnetic field of magnet I N S MOVE MAGNET
The induced emf always opposes the change in flux Faraday’s Law / Lenz’s Law The induced emf always opposes the change in flux transformer emf Move magnet away from loop B decreases in loop Flux decreases in loop Current induced Current produces magnetic field in loop This magnetic field is in same direction to magnetic field of magnet I N S MOVE MAGNET
Faraday’s Law N S MOVE MAGNET Suppose loop is stationary so we have only transformer emf. Equivalent battery to drive current around loop N S MOVE MAGNET
Faraday’s Law in derivative form: Valid for all points in space
The induced emf always opposes the change in Note on: Faraday’s Law / Lenz’s Law The induced emf always opposes the change in flux WE KNOW THAT 𝐵 𝑡 ∝𝐼 𝑡 𝐵 𝑡 = 𝐾 ∗𝐼 𝑡 constant Inductance: relates the induced emf to the time rate of change of the current 𝑣 𝑡 =−𝐿 𝜕𝐼 𝑡 𝜕𝑡
Note on: Faraday’s Law / Lenz’s Law Rotating current loop in constant magnetic field 𝜔 O 𝜃 𝐵 X 𝐵 ∙ 𝑑𝑎 =𝐵 𝑑𝑎 𝑐𝑜𝑠 𝜔𝑡 Sinusoidal voltage change Amplitude of voltage depends on B, A and POWER GENERATOR
? Note on: Faraday’s Law / Lenz’s Law Moving rod in constant magnetic field 𝐵 X X X X X X X X L X X X X - + Which end of the rod is positive and which end is negative? ? + -
ELEC 3105 Basic EM and Power Engineering Displacement Current
NOW it might make sense, if Figure 8.1 Application of Ampere’s law to a circuit with an air-filled capacitor and time-varying current. NOW it might make sense, if i = dD/dt Sp Taken from ELEC 3909
Displacement Current For surface A chosen So far we have the following expression for Ampere’s law. WIRE Certainly there is no problem in evaluating the integral. The path shown encloses an area A through which the wire cuts. We are in fact in the process of charging the capacitor through the current I of the wire. Capacitor For surface A chosen
Displacement Current For path A’ chosen Capacitor So far we have the following expression for Ampere’s law. WIRE Certainly there is a problem in evaluating the integral. The path shown encloses an area A’ through which the wire does not cut. Yet the integral value evaluated here for the surface that does not contain the wire A’ and the surface of the previous slide A which does contain the wire must be the same since the surfaces A’ and A are arbitrary. We are in fact in the process of charging the capacitor through the current I of the wire. Capacitor For path A’ chosen Both integrals should give the same result??????
Displacement Current We have a problem Capacitor Capacitor WIRE WIRE For different surfaces
Solution to the problem Displacement Current Solution to the problem Between the capacitor plates we have a changing electric flux density. This changing electric flux density is equivalent to a current density.
Solution to the problem Displacement Current Solution to the problem Ampere’s law in integral form Apply Stoke’s theorem Ampere’s law in derivative form
Ampere’s Law in derivative form: Valid for all points in space Displacement Current