 # Electromagnetic Induction

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Electromagnetic Induction
Chapter 31 (cont.) Faraday’s Law & Lenz’s Law Induced Electric Fields Mutual Inductance Self-Inductance Inductors in Circuits Magnetic Energy 1

A changing magnetic field induces an emf proportional to the rate of change of magnetic flux: e = - dB / dt The emf is about some loop – and the flux is through that loop. Lenz’s law: the induced emf is directed so that any induced current opposes the change in magnetic flux that causes the induced emf.

Rotating Loop - The Electric Generator
B q Consider a loop of area A in a uniform magnetic field B. Rotate the loop with an angular frequency w . The flux changes because angle q changes with time: q = wt. Hence dFB/dt = d(B · A)/dt = d(BA cos q)/dt = BA d(cos(wt))/dt = - BAw sin(wt) q A

Rotating Loop - The Electric Generator
B A q dFB/dt = - BAw sin(wt) Then by Faraday’s Law this motion causes an emf e = - dFB /dt = BAw sin(wt) This is an AC (alternating current) generator.

Induced Electric Fields
Consider a stationary wire in a time-varying magnetic field. A current starts to flow. x dB/dt So the electrons must feel a force F. It is not F = qvxB, because the charges started stationary. Instead it must be the force F=qE due to an induced electric field E. That is: A time-varying magnetic field B causes an electric field E to appear!

Induced Electric Fields
Consider a stationary wire in a time-varying magnetic field. A current starts to flow. x dB/dt So the electrons must feel a force F. It is not F = qvxB, because the charges started stationary. Instead it must be the force F=qE due to an induced electric field E. Moreover E around the loop gives a voltage diff DV=E·dl. And this must be the emf:  =  E·dl o

Induced Electric Fields
This gives another way to write Faraday’s Law:  E·dl = - dB/dt o A technical detail: The electrostatic field E is conservative:  E·dl = 0. Consequently we can write E = - V. The induced electric field E is NOT a conservative field. We can NOT write E = -V for an induced field. o

Work or energy difference Work or energy difference
Electrostatic Field Induced Electric Field F = q E F = q E Vab = -  E·dl o  E · dl = - dB/dt  E·dl  0  E·dl = 0 and Ee = V o o Conservative Nonconservative Work or energy difference does NOT depend on path Work or energy difference DOES depend on path Caused by stationary charges Caused by changing magnetic fields

Induced Electric Fields
 E · dl = - dB/dt o x B Faraday’s Law Now suppose there is no conductor: Is there still an electric field? YES! The field does not depend on the presence of the conductor. E For a dB/dt with axial or cylindrical symmetry, the field lines of E are circles. dB/dt

Mutual Inductance Bof 1 through 2 I 1 2
Two coils, 1 & 2, are arranged such that flux from one passes through the other. We already know that changing the current in 1 changes the flux (in the other) and so induces an emf in 2. This is known as mutual inductance. I Bof 1 through 2 1 2 2

Mutual Inductance I 1 2 Bof 1 through 2 The mutual inductance M is the proportionality constant between F2 and I1: F2 = M I1 so dF2 /dt = M dI1 /dt Then by Faraday’s law: 2= - dF2 /dt = - M dI1 /dt Hence M is also the proportionality constant between 2 and dI1 /dt. 3

Mutual Inductance 1 H = 1Weber/Amp = 1 V-s/A
M arises from the way flux from one coil passes through the other: that is from the geometry and arrangement of the coils. Mutual means mutual. Note there is no subscript on M: the effect of 2 on 1 is identical to the effect of 1 on 2. The unit of inductance is the Henry (H). 1 H = 1Weber/Amp = 1 V-s/A 4

Self Inductance A changing current in a coil can induce an emf
in itself…. I B If the current is steady, the coil acts like an ordinary piece of wire. But if the current changes, B changes and so then does FB, and Faraday tells us there will be an induced emf. Lenz’s law tells us that the induced emf must be in such a direction as to produce a current which makes a magnetic field opposing the change.

Self Inductance The self inductance of a circuit element (a coil, wire, resistor or whatever) is L = FB/I. Then exactly as with mutual inductance  = - L dI/dt. Since this emf opposes changes in the current (in the component) it is often called the “back emf”. From now on “inductance” means self-inductance. 6

Example: Finding Inductance
What is the (self) inductance of a solenoid with area A, length d, and n turns per unit length? In the solenoid B = m0nI, so the flux through one turn is fB = BA = m0nIA The total flux in the solenoid is (nd)fB Therefore, FB = m0n2IAd and so L = FB/I gives L = m0n2Ad (only geometry) 16

Inductance Affects Circuits and Stores Energy
This has important implications….. First an observation: Since  cannot be infinite neither can dI/dt. Therefore, current cannot change instantaneously. We will see that inductance in a circuit affects current in somewhat the same way that capacitance in a circuit affects voltage. A ‘thing’ (a component) with inductance in a circuit is called an inductor. 7

RL Circuit e0 While the switch is open current can’t flow.
We start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L

RL Circuit e0 e0 eL While the switch is open current can’t flow.
We start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L When the switch is closed current I flows, growing gradually, and a ‘back emf’ eL=- L dI/dt is generated in the inductor. This opposes the current I. + - S I e0 eL

RL Circuit e0 e0 eL e0 While the switch is open current can’t flow.
We start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L When the switch is closed current I flows, growing gradually, and a ‘back emf’ eL=- L dI/dt is generated in the inductor. This opposes the current I. + - S I e0 eL S + e0 After a long time the current becomes steady. Then eL is zero. Is -

RL Circuit e0/R e0 I t/(L/R) When the switch S is closed S + I -
the current I increases exponentially from I = 0 to I = e0/R e0/R I 1 2 3 4 5 t/(L/R)

RL Circuit Analysis e0 e0 - IR - LdI/dt = 0 Use the loop method R S +
LdI/dt = e0 - IR = -R[I-(e0 /R)] dI / [I-(e0 /R)] = - (R/L) dt   dI / (I-(e0 /R)) = - (L/R)  dt ln[I-(e0 /R)] – ln[-(e0 /R)] = - t / (L/R)  ln[-I/(e0 /R) + 1] = - t / (L/R) -I/(e0 /R) + 1 = exp (- t / (L/R))  I/(e0 /R) = 1 - exp (- t / (L/R)) I = (e0 /R) [1 - exp (- t / (L/R))] 9

with time constant  = L / R
RC Circuit Analysis R + - S I e0 L I = (e0/R) [1 - exp (- t / (L/R))] The current increases exponentially with time constant  = L / R e0 /R I t = 0  I = 0 t =   I = e0 /R 1 2 3 4 5 t/(L/R) 9

eL = L (e0/R) (-R/L) exp (- t / (L/R))
Inductor’s emf eL R + - S I e0 L eL = - L (dI/dt) I = (e0/R) [1 - exp (- t / (L/R))] eL = L (e0/R) (-R/L) exp (- t / (L/R)) eL 1 2 3 4 5 t/(L/R) -e0 eL = - e0 exp (- t / (L/R)) t = 0  eL = - e0 t =   eL = 0 9

Decay of an RL Circuit S R + e0 - L After I reaches e0/R move the switch as shown. The loop method gives eL - IR = 0 or eL = IR Remember that eL = -L dI/dt  -L dI/dt = IR dI/I = - dt / (L/R)   dI/I = -  dt / (L/R) ln I/I0 = - t / (L/R)  I = I0 exp [- t / (L/R)] But I0 = e0/R Then: I = (e0/R) exp [- t / (L/R)] 12

Inductors in Circuits The presence of inductance prevents currents from changing instantaneously. The time constant of an RL circuit is  = L/R. When the current is not flowing the inductor tries to keep it from flowing. When the current is flowing the inductor tries to keep it going. 13

Energy Stored in an Inductor
+ - S I e0 L Recall the original circuit when current was changing (building up). The loop method gave: e0 - IR + eL = 0 Multiply by I and use eL = - L dI/dt Then: Ie0 - I2R - ILdI/dt = 0 or: Ie0 - I2R – d[(1/2)LI2]/dt = 0 {d[(1/2)LI2]/dt=ILdI/dt} 14

Think about Ie0 - I2R - d((1/2)LI2)/dt = 0
Ie0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. 15

Think about Ie0 - I2R - d((1/2)LI2)/dt = 0
Ie0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. Hence we’d like to interpret d((1/2)LI2)/dt as the rate at which energy is stored in the inductor. In creating the magnetic field in the inductor we are storing energy 15

Think about Ie0 - I2R - d((1/2)LI2)/dt = 0
Ie0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. Hence, we’d like to interpret d[(1/2)LI2]/dt as the rate at which energy is stored in the inductor. In creating the magnetic field in the inductor we are storing energy The amount of energy in the magnetic field is: UB = (1/2) LI2 15

Energy Density in a Magnetic Field
We have shown Apply this to a solenoid: Dividing by the volume of the solenoid, the stored energy density is: This turns out to be the energy density in a magnetic field uB = B2/(20) 17

Summary We defined mutual and self inductance,
Calculated the inductance of a solenoid. Saw the effect of inductance in RL circuits. Developed an expression for the stored energy. Derived an expression for the energy density of a magnetic field. Next class we will start learning about alternating-current (AC) circuits, containing resistors, capacitors, and inductors. 18