Faraday s Law of Induction

Faraday s Law of Induction Potential drop along the closed contour is minus the rate of change of magnetic flu. We can change the magnetic flu in several ways including changing the magnitude of the magnetic field, changing the area of the loop, or by changing the angle the loop with respect to the magnetic field 10

Lenz s Law (1) Lenz s Law defines a rule for determining the direction of an induced current in a loop An induced current will have a direction such that the magnetic field due to the induced current opposes the change in the magnetic flu that induces the current Meaning, if the flu increases, the induced B-field will be directed against eternal B-field 25

Lenz s Law - Induced current - Induced magnetic field in the direction so to oppose the change 3

Lenz s Law - Induced current - Induced magnetic field in the direction so to oppose the change i 3

Lenz s Law - Induced current - Induced magnetic field in the direction so to oppose the change E i 3

Eddy currents E-field line is circular. If within a conductor, it will drive current: eddy current. That current will produce its own B-field, opposing the eternal change (Lenz s law) 4

Eddy currents E-field line is circular. If within a conductor, it will drive current: eddy current. That current will produce its own B-field, opposing the eternal change (Lenz s law) E 4

Eddy currents E-field line is circular. If within a conductor, it will drive current: eddy current. That current will produce its own B-field, opposing the eternal change (Lenz s law) E i 4

Inductive E-field: work along closed path 5

Inductive E-field: work along closed path v 5

Inductive E-field: work along closed path v E 5

Inductive E-field: work along closed path v E-field lines are closed! This is very different from E-field from a point charge! E 5

demo: magnet falling through pipe The B-field induced by eddy currents will oppose the effect that induced the current = motion of the magnet. There will be breaking force. B-field generated by eddy currents will cause a repulsive or drag force between the conductor and the eternal magnet: magnetic breaking. gravitational energy is dissipated by resistivity, can heat an object. 6

Demos: magnet in a tube Eddy currents Jumping ring 7

Inductance i The unit of inductance is the henry (H) given by 8

Inductance i B The unit of inductance is the henry (H) given by 8

Inductance Magnetic flu through the contour is proportional to current, Φ i i B The unit of inductance is the henry (H) given by 8

Inductance Magnetic flu through the contour is proportional to current, Φ i Φ=Li i B The unit of inductance is the henry (H) given by 8

Inductance Magnetic flu through the contour is proportional to current, Φ i Φ=Li L - inductance (self-inductance) i B The unit of inductance is the henry (H) given by 8

Inductance Magnetic flu through the contour is proportional to current, Φ i Φ=Li L - inductance (self-inductance) i compare with capacitance: q= C V The unit of inductance is the henry (H) given by B 8

Inductance of a current loop. 9

B-field of a ring current B = µ0 4π I dl r r 3 db z = db cos β = µ 0 4π idl r 2 cos β cos β = R/r r = z 2 + R 2 B = µ 0 4π 2πR 2 i (z 2 + R 2 ) 3/2 Recall: E-field of a dipole ~ 1/z 3, similarly, B ~ 1/z 3 10

Self-inductance of a current loop. In the plane of the loop Φ R 2 B µ 0 ir B µ 0 2 i R Φ=Li L µ 0 R Similar to capacitance, inductance is a geometrical property 11

Inductance (1) Consider a long solenoid with N turns carrying a current i Same current, flu adds: large inductance. This current creates a magnetic field in the center of the solenoid resulting in a magnetic flu of Φ B The quantity NΦ B, called the flu linkage, is always proportional to the current with a proportionality constant called the inductance L 41

Inductance (1) Consider a long solenoid with N turns carrying a current i Same current, flu adds: large inductance. This current creates a magnetic field in the center of the solenoid resulting in a magnetic flu of Φ B Φ=Li The quantity NΦ B, called the flu linkage, is always proportional to the current with a proportionality constant called the inductance L 41

Inductance of a Solenoid Consider a solenoid with cross sectional area A and length l The flu linkage is n is the number of turns per unit length and B = µ 0 in The inductance of a solenoid is then The inductance of a solenoid depends only on its geometry 43

Self Inductance and Mutual Induction Consider the situation in which two coils, or inductors, are close to each other A current in the first coil produces magnetic flu in the second coil Changing the current in the first coil will induce an emf in the second coil However, the changing current in the first coil also induces an emf in itself This phenomenon is called self-induction The resulting emf is termed the self-induced emf 44

Self Induction Faraday s Law of Induction tells us that the self-induced emf for any inductor is given by Thus in any inductor, a self-induced emf appears when the current changes with time This self-induced emf depends on the time rate change of the current and the inductance of the device Lenz s Law provides the direction of the self-induced emf The minus sign epresses that the induced emf always opposes any change in current 45

Self Inductance: Increasing Current In the figure below, the current flowing through an inductor is increasing with time Thus a self-induced emf arises to oppose the increase in current 46

Self Inductance: Decreasing Current In the figure below, the current flowing through an inductor is decreasing with time Thus a self-induced emf arises to oppose the decrease in current 47

Mutual inductance Φ 2 = L 12 i 1 Φ 1 = L 21 i 2 L 21 = L 12 18

Demo: Faraday s law 19

Energy of a Magnetic Field The instantaneous power provided by the emf source is the product of the current and voltage in the circuit Integrating this power over the time it takes to reach a final current yields the energy stored in the magnetic field of the inductor 59

Energy of B-field L = µ 0 n 2 V W = Li2 2 = µ 0 2 n2 i 2 V ni = B/µ 0 W = B2 2µ 0 V w = B2 2µ 0 Energy density of B-field 21

RL Circuits (1) We have assumed that our inductors have no resistance Now let s treat inductors that have resistance Reminder: RC circuits We know that if we place a source of eternal voltage, V emf, into a single loop circuit containing a resistor R and a capacitor C, the charge q on the capacitor builds up over time as where the time constant of the circuit is given by τ C = RC The same time constant governs the decrease of the initial charge q in the circuit if the emf is suddenly removed 48

RL Circuits (2) If we place an emf in a single loop circuit containing a resistance R and an inductor L, a similar phenomenon occurs If we had connected only the resistor and not the inductor, the current would instantaneously rise to the value given by Ohm s Law as soon as we closed the switch However, in the circuit with both the resistor and the inductor, the increasing current flowing through the inductor creates a self-induced emf that tends to oppose the increase in current 49

RL Circuits (3) As time passes, the change in current decreases and the opposing self-induced emf decreases and after a long time, the current is steady We can use Kirchhoff s loop rule to analyze this circuit assuming that the current i at any given time is flowing through the circuit in a counterclockwise direction The emf source represents a gain in potential, +V emf, and the resistor represents a drop in potential, -ir 50

RL Circuits (4) The self-inductance of the inductor represents a drop in potential because it is opposing the increase in current The drop in potential due to the inductor is proportional to the time rate change of the current and is given by 51

RL Circuits (5) Thus we can write the sum of the potential drops around the circuit as We can rewrite this equation as The solution to this differential equation is We can see that the time constant of this circuit is τ L = L/R 52

RL Circuits (6) Now consider the case in which an emf source had been connected to the circuit and is suddenly removed We can use our previous equation with V emf = 0 to describe the time dependence of this circuit 53

RL Circuits (7) The solution to this differential equation is where the initial conditions when the emf was connected can be used to determine the initial current, i 0 = V emf /R This equation describes a single loop circuit with a resistor and an inductor that initially has a current i 0 The current drops with time eponentially with a time constant τ L = L/R and after a long time the current in the circuit is zero 54

RL Circuits (8) Variation with time of (a) the voltage across the resistor in a RL circuit and (b) the potential difference across the inductor (R = 2000 Ω, L = 4 H, and V emf = 10 V). V emf resistor V emf inductor 55

Eample: RL Circuits (1) 56

Eample: RL Circuits (1) A solenoid has an inductance of 53 mh and a resistance of 0.37 Ω. Question: If the solenoid is connected to a battery, how long will the current take to reach half its final equilibrium value? 56

Eample: RL Circuits (1) A solenoid has an inductance of 53 mh and a resistance of 0.37 Ω. Question: If the solenoid is connected to a battery, how long will the current take to reach half its final equilibrium value? Answer: 56

Eample: RL Circuits (1) A solenoid has an inductance of 53 mh and a resistance of 0.37 Ω. Question: If the solenoid is connected to a battery, how long will the current take to reach half its final equilibrium value? Answer: We can mentally separate the solenoid into a resistance and an inductance that are wired in series with a battery 56

Eample: RL Circuits (1) A solenoid has an inductance of 53 mh and a resistance of 0.37 Ω. Question: If the solenoid is connected to a battery, how long will the current take to reach half its final equilibrium value? Answer: We can mentally separate the solenoid into a resistance and an inductance that are wired in series with a battery Step 1: Kirchhoff s loop rule: 56

Eample: RL Circuits (2) 57

Eample: RL Circuits (2) Step2: The current i increases eponentially from zero to its final equilibrium value of V emf /R. Let t 0 be the time that the current i takes to reach half the equilibrium value: 57