Chapter

Faraday’s Law and Inductance

Blinn College - Physics 1402 - Terry Honan

Faraday's law describes how a generator works.  A changing magnetic flux through a conducting loop induces an EMF (Electromotive Force) around the loop.   Faraday's law will become our fourth of Maxwell's equations.

. - Faraday’s Law

In the previous chapter we introduced magnetic flux in our brief discussion of Gauss’s law for magnetism. The flux is (Subscript[Φ, m])=∑ (Subscript[B, ⊥])A. Now we will omit the m subscript.

Φ=∑ (Subscript[B, ⊥])A

IF we have a flat surface and a uniform magnetic field then the flux becomes

Φ=(Subscript[B, ⊥])A= B A cos θ

where we defined θ as the angle between the normal to a surface and the magnetic field.

We can demonstrate magnetic induction by moving a magnet toward a solenoid connected to a galvanometer, where a deflection of the galvanometer's needle indicates a current.  Moving the magnet toward the solenoid induces an EMF in the solenoid, which creates a current in the solenoid-galvanometer circuit.  Pulling the magnet away from the solenoid induces a current in the opposite direction.  If the magnet is at rest in the solenoid no current is induced.  Moreover, increasing the number of magnets increases these effects.  It is clear that the induced EMF depends on the change in the flux.  If (Subscript[ℰ, ave]) is the average induced EMF and Φ is the magnetic flux through each loop of the solenoid then we get the proportionality

(Subscript[ℰ, ave])∝ΔΦ.

If the speed of the magnet is increased the effect is enhanced and slowing it diminishes it.  This suggests an inverse proportionality with the time.

(Subscript[ℰ, ave])∝(1/(Δt)).

Since in a coil all the loops are connected in series it follows that  (Subscript[ℰ, ave]) is proportional to the number of loops N.

Combining these proportionalities gives

(Subscript[ℰ, ave])∝N ((ΔΦ)/(Δt))

and it turns out that the proportionality becomes an equality when we consider the magnitude of the average induced EMF |(Subscript[ℰ, ave])| and the absolute value of the change in flux with time

(Subscript[ℰ, ave])=N |((ΔΦ)/(Δt))|.

If we consider the polarity (sign) of the induced EMF then this adds a sign, which is known as Lenz's law.

(Subscript[ℰ, ave])=-N ((ΔΦ)/(Δt)).

We will discuss Lenz's law in detail later.

Letting Δt go to zero these expressions become

|ℰ|=N |((ΔΦ)/(Δ t))|    and   ℰ=-N ((ΔΦ)/(Δ t))  (small Δ t)

This is analogous to the definitions of average and instantaneous velocity in kinematics where  (Subscript[v, ave])=((Δx)/(Δt)) and v=((Δx)/(Δt))  (small Δ t).

. - Motional EMF

If a conductor moves in a magnetic field there is a magnetic force on the charge carriers.  This magnetic force does work on the charge carriers.  The EMF is the work per charge.

ℰ=(W/q).

Translation of a Conducting Rod in a Uniform Field

Consider a conducting rod of length ℓ translating in a uniform magnetic field.  Take the rod, its velocity and the field to be mutually perpendicular.

If q is some charge carrier then the force on it is

F=q v (Subscript[B, ⊥]) ⟹  F=q v B

The work is

W=F Δ (Subscript[r, ∥])=F ℓ=q v B ℓ.

Using  ℰ=W/q  gives the induced EMF across the rod

ℰ=v B ℓ

How can one measure this EMF?  If a voltmeter is connected across the ends of the rod and it is moved with it then the same EMF is induced in the leads to the meter and it reads zero.  To avoid this, imagine the rod moving with its ends sliding along a conducting rail.  If the voltmeter is connected between the rails then it would read this voltage.

A DC Generator and Conservation of Energy

In fact, what we have here is a simple DC generator.  Instead of connecting a voltmeter between the rails we could connect anything and it would be given a steady DC voltage.  If a DC motor is connected then this motor could do work.  We must address the question of conservation of energy.  Where does this energy come from?

To see this place a load resistor R across the conducting rails.  Ohm's law gives the current through the load

I=(ℰ/R).

The rate of power dissipation in the load, which is the power output of the generator, is

(Subscript[𝒫, out])=I ℰ=I v B ℓ,

where we are making the idealizing assumption that all the resistance in the circuit is in the load.

This is a complete circuit, so all the current through the load passes through the rod.  A current through a conductor in a magnetic field creates a backward magnetic force,  (Subscript[(OverVector[F]), mag]).

(Subscript[F, mag])=I ℓ (Subscript[B, ⊥])  ⟹  (Subscript[F, mag])=I ℓ B

To keep the rod moving at a constant speed there must be zero net force, so there must be some forward external applied force (Subscript[(OverVector[F]), app]).  Making another idealizing assumption of no friction we get

(Subscript[(OverVector[F]), app])=-(Subscript[(OverVector[F]), mag])  ⟹  (Subscript[F, app])=(Subscript[F, mag])=I ℓ B

The applied force does work and this is the source of the energy.  The rate that it does work is the power (Subscript[𝒫, in]).  Power is related to force by

𝒫=((ΔWork)/(Δ t))=((F Δ (Subscript[r, ∥]))/(Δ t))=F (Subscript[v, ∥])   (small Δ t)

so we get in this case

(Subscript[𝒫, in])=(Subscript[F, app]) v=I ℓ B v.

Comparing this with the power output gives

(Subscript[𝒫, in])=(Subscript[𝒫, out]).

If we relax our idealizing assumptions and include mechanical friction and resistance in the circuit elsewhere than the load then we get

(Subscript[𝒫, in])>(Subscript[𝒫, out]).

Example . - Motional EMF

Solution

ℰ=v B ℓ=0.324(QuantityUnit)

Solution

Solution

V=ℰ=I R   ⟹   R=(ℰ/I)=58.9(QuantityUnit)

Solution

Solution

𝒫=(Subscript[𝒫, out])=I ℰ=1.78×(10^(-3))(QuantityUnit)

. - Faraday’s Law Examples

Usually when we use Faraday's law we will only consider the magnitude of the induced EMF or current.  We will see how to find the polarity in the Lenz's law discussion that follows.  We can neglect polarity information by inserting absolute values into Faraday's law.

|(Subscript[ℰ, ave])|=N |((ΔΦ)/(Δt))|  and   |ℰ|=|(Subscript[ℰ, ave])|     (small Δ t)

If one is given some problem involving induced currents then this is related to the induced EMF by Ohm's law

ℰ=I R

The Translating Rod

There is an equivalence between motional EMF and Faraday's law.  For the case of the translating rod that we analyzed above using motional EMF, we ought to be able to derive the same results using Faraday's law directly.  If a load resistor is placed in the circuit then the complete circuit consists of the rod, rails and load.  As the rod moves the area inside the circuit loop is increasing and so is the flux.  In the time Δ t the rod moves by v Δ t and the area increases by

Δ A=ℓ v Δ t

The flux, since the field is uniform and perpendicular to the surface, is Φ=B A and since the field is constant we get ((Δ Φ)/(Δ t))=B ((Δ A)/(Δ t)). The induced EMF becomes

|ℰ|=|((Δ Φ)/(Δ t))|=B ((Δ A)/(Δ t))=B ℓ v

which is the same as the expression derived using motional EMF considerations.

The AC Generator

Consider a solenoid or coil with its central axis rotating with angular velocity ω in a uniform magnetic field of magnitude B.  Take the central axis to be perpendicular to the rotational axis and  the rotational axis to be perpendicular to the field.  The coil has N turns and each loop has a cross-sectional area A. Take the angle between the axis of the coil and the field to be θ.  This varies by

θ=ω t

The flux through a loop is

Φ=B A cos θ=B A cos ω t.

By Faraday's law  ℰ=-N ((Δ Φ)/(Δ t)) (for small Δ t) we get the induced EMF as a function of time to be

ℰ(t)=N B A ω sin ω t.

This is an AC voltage.  We will discuss AC in detail in the AC circuit chapter.  We will state for future reference that the peak EMF is

(Subscript[ℰ, max])=N B A ω  where  ω=2π f

and the frequency of AC is the same as the rotational frequency. We will see in the next chapter that the angular frequency ω is related to the frequency f by ω=2π f.

The AC Motor

With a generator one does work to turn the coil and that energy is converted to electrical energy. A generator run in reverse is a motor and electrical energy is converted to work.

Example . - Faraday’s Law

. - Lenz’s Law

The minus sign in Faraday's law is known as Lenz's law.  It gives the polarity of the induced EMF.  If the circuit is completed with a load resistance the induced EMF causes an induced current.  The induced current induces a magnetic field (Subscript[(OverVector[B]), ind]) which in turn induces a flux (Subscript[Φ, ind]) through the loop.

The induced flux tends to cancel the change in  the flux.

This cancellation is exact for a superconductor.

The magnetic flux  through a superconducting loop is a constant.

Using Lenz's Law

A simple procedure for applying Lenz's law follows. Identify a sign convention and then make a table consisting of the appropriate signs for Φ, ((ΔΦ)/(Δ t)), (Subscript[Φ, ind]) and give the sense of circulation for ℰ.  Flux is a scalar but it has a sign; we can associate the signs of our flux variables with the directions of the normals to the loop. The first step is to identify the circuit loop and choose a positive direction relative to the normal to the loop. Make a table with the first three columns having signs and the fourth having a sense of rotation.

The sign of Φ is the direction of the field through the loop. If the field is in the same direction as the normal it is positive and negative otherwise.

The sign of Δ Φ/Δ t is the same as Φ when the flux is increasing, and opposite to Φ when the flux is decreasing.

By Lenz's law, the sign of (Subscript[Φ, ind]) is always opposite to that of ((Δ Φ)/(Δ t)).

To get the direction of the induced emf and current, ℰ and I, use the right-hand rule to determine which sense current gives the induced flux.

Example . - Polarity of Motional EMF

Solution

({{Positive, Sign of, Sign of, Sign of, Sense of}, {normal, Φ, dΦ/dt, (Subscript[Φ, ind]), ℰ or I}, {, -, -, +, }})

Example . - Magnets Falling through Conducting Loops

Solution

({{Positive,  , Sign of, Sign of, Sign of, Sense of}, {normal,  , Φ, ΔΦ/Δ t, (Subscript[Φ, ind]), ℰ or I}, {

, Enters, -, -, +,   counterclockwise}, {, Leaves, -, +, -,         clockwise         }})

Solution

Solution

({{Positive,  , Sign of, Sign of, Sign of, Sense of}, {normal,  , Φ, ΔΦ/Δ t, (Subscript[Φ, ind]), ℰ or I}, {

, Enters, -, -, +,   counterclockwise}, {, Leaves, +, -, +,   counterclockwise}})

Example . - Induced Current around an Inner Loop Due to a Changing Current in an Outer Loop

Solution

({{Positive,  , Sign of, Sign of, Sign of, Sense of}, {normal,  , Φ, ΔΦ/Δ t, (Subscript[Φ, ind]), ℰ or I}, {

, Closed, +, +, -,         clockwise           ⟹   Current to the right through R}, {, Opened, +, -, +,   counterclockwise   ⟹   Current to the left through R  }})

. - Inductance

(Self) Inductance

As a consequence of Faraday's law a changing current through one coil induces an EMF in another coil; this is known as mutual inductance.  Similarly, a changing flux in a coil induces an EMF in the same coil; this is self inductance and a circuit component with inductance is called an inductor.

A changing current through a coil can induce an EMF (voltage) across the coil. This is called self inductance.  When we use the term inductance by itself self inductance is implied.  A current through a coil creates a field and that causes a flux through the coil itself.

I   ⟹   (OverVector[B])   ⟹   Φ

A changing current creates a changing flux with induces an EMF  in the coil.

((Δ I)/(Δ t))   ⟹   ((Δ Φ)/(Δ t))   ⟹   ℰ

The above relationships are proportionalities.  The constant of proportionality is defined as the inductance L.

ℰ=-L ((Δ I)/(Δ t))   (small Δ t)

The sign in the above expression is due to Lenz's law.  Take ΔV to be the change in the voltage when moving through the inductor in the direction of the current.  A simple Lenz's law analysis shows that if the current is increasing the voltage change is negative.  If we write V as the voltage drop we get

ΔV=-L ((Δ I)/(Δ t))  and  V=L ((Δ I)/(Δ t)).

The sign conventions for inductors is the same as that for resistors and capacitors, V is the voltage drop when moving with the current.

ΔV=-I R  and  V=I R

ΔV=- (Q/C)  and  V=(Q/C)  where  I=((Δ Q)/(Δ t)) (for small Δ t)

Units:  The SI unit for Inductance is:  henry=(QuantityUnit)

Inductance of a Long Solenoid

Consider a long solenoid of length ℓ, with N turns and a cross-sectional area A.  As before we define n as the number of turns per length  n=N/ℓ.  Passing from the current to the field to the flux gives

I   ⟹   B=(Subscript[μ, 0]) n I   ⟹   Φ=B A=(Subscript[μ, 0]) n I A

and using Faraday's law we get

ℰ=-N ((Δ Φ)/(Δ t))=-N (Subscript[μ, 0]) n A ((Δ I)/(Δ t))    (for smallΔ t).

Using  ℰ=-L Δ I/Δ t (for small Δ t), we can read the inductance from this expression

L=(Subscript[μ, 0]) ((N^2)/ℓ) A,

where we used n=N/ℓ.

Example . - A Long Solenoid as an Inductor

Solution

ℓ=0.75(QuantityUnit), N=200, (Subscript[μ, 0])=4π×(10^(-7))(((QuantityUnit))/(((QuantityUnit))^2))  and  r=0.030(QuantityUnit)  ⟹  A=π (r^2)=0.0028274(((QuantityUnit))^2)

L=(Subscript[μ, 0]) ((N^2)/ℓ) A=1.8950×(10^(-4))(QuantityUnit)

ℰ=-L ((Δ I)/(Δ t))   (small Δ t)   ⟹  (Subscript[ℰ, ave])=-L ((Δ I)/(Δ t))

Δ I=11(QuantityUnit)-25(QuantityUnit)=-14(QuantityUnit)   and   Δ t=3.5(QuantityUnit)=3.5×(10^(-3))(QuantityUnit)

|(Subscript[ℰ, ave])|=L |((Δ I)/(Δ t))| =0.758(QuantityUnit)

. - Energy Considerations

Energy in an Inductor

Inductors, like capacitors, store energy, while resistors dissipate energy.  Use U to denote the energy in an inductor.  The rate that energy is being stored in an inductor is (for small Δ t)

((Δ U)/(Δ t))=𝒫=I V=I L ((Δ I)/(Δ t))   ⟹   Δ U=I L Δ I

Increase the current from 0 to I; call the intermediate current i and then we have 0≤i≤I and Δ U=L i Δ i . Summing over these small Δ U values as i varies from 0 to I gives the total energy.

U=L ∑ i Δ i

The diagram above shows that ∑ i Δ i=(1/2)(I^2). We then get the expression for the energy in an inductor.

U=(1/2)L (I^2)

Example . - A Long Solenoid as an Inductor (Continued)

Solution

U=(1/2)L (I^2)=0.0307(QuantityUnit)

Energy in a Magnetic Field

In the capacitance chapter we derived an expression for the energy density (Energy/Volume) in an electric field.

u=(1/2)(Subscript[ε, 0]) (E^2)

To derive this we used the fact that the electric field is uniform inside a parallel plate capacitor.  Combining expressions for the energy in a capacitor and for the capacitance gave the above expression for u.  A similar analysis will give the energy density in a magnetic field.

The magnetic field is uniform inside a long solenoid.  Combining the expression for the inductance of a long solenoid with the energy in an inductor gives

U=(1/2)L (I^2)  and  L=(Subscript[μ, 0]) (n^2) A ℓ   ⟹   U=(1/2)(Subscript[μ, 0]) (n^2) A ℓ (I^2)

Using B=(Subscript[μ, 0]) n I and u=U/Volume=U/(A ℓ) gives the energy density in a magnetic field.

u=(1/(2 (Subscript[μ, 0]))) (B^2)

Example . - Energy Density

. - RL Circuits

Decaying Current

Begin with switch A closed and B opened.  This creates a current through the inductor and resistor. Close switch B and then open A.  This causes the current to flow through the top branch of the above circuit.  Applying the loop rule around the circuit gives

0=L ((Δ I)/(Δ t))+R I.

Rewrite this as

((Δ I)/(Δ t))=-(1/τ) I,

where the time constant τ is defined by

τ=(L/R).

Take the initial current to be (Subscript[I, 0]). Solving the equation for t is a calculus problem. We will just write down the solution.

I(t)=(Subscript[I, 0]) (e^(-t/τ)),

This is a simple exponential decay, analogous to the decay of the charge for a discharging capacitor.

The initial energy in the inductor U=(1/2)L  (Subscript[I, 0]^2) is converted to heat in the resistor.

Growing Current

With both switches opened giving zero current, close switch A at t=0.  This causes the current to gradually build up to a steady-state value.  Apply the loop rule to the circuit gives the first order ODE.

ℰ=L ((Δ I)/(Δ t))+R I.

Using the same time constant τ we can write down the solution where we take the initial current, at time zero, to be zero.

I(t)=(Subscript[I, ∞]) (1-(e^(-t/τ)))  where  (Subscript[I, ∞])=(ℰ/R)

is the steady-state current.

Example . - A Long Solenoid as an Inductor (Continued... More)

Solution

L=1.8950×(10^(-4))(QuantityUnit), ℰ=12(QuantityUnit) and  R=0.85(QuantityUnit)

(Subscript[I, ∞])=ℰ/R=14.1(QuantityUnit)

Solution

τ=L/R=0.00022294(QuantityUnit)=0.22294(QuantityUnit)

I(t)=(Subscript[I, ∞]) (1-(e^(-t/τ))) =10.4(QuantityUnit)

Solution

I(t)=(Subscript[I, 0]) (e^(-t/τ)) =3.68(QuantityUnit)

. - The Transformer

In addition to self inductance where a changing current through a coil induces an EMF across its terminals, there is mutual inductance where a changing current in one coil induces an EMF in another coil. We will consider only one special case of mutual inductance, the transformer.

A transformer is a case of mutual inductance where, ideally, all of the flux from one coil passes through the other.  Take the primary coil to have (Subscript[N, 1]) turns and the secondary to have (Subscript[N, 2]).  This can be achieved to a good approximation by wrapping both coils around the same soft ferromagnetic core.   The core amplifies the field due to the current and directs the field lines around the loop of the core

The fluxes are assumed to be equal.

Φ=(Subscript[Φ, 1])=(Subscript[Φ, 2])

Then by Faraday's law

(Subscript[V, 1])=-(Subscript[N, 1])((Δ Φ)/(Δ t))  and  (Subscript[V, 2])=-(Subscript[N, 2])((Δ Φ)/(Δ t))

we get a proportionality between the voltage and number of turns.

((Subscript[V, 2])/(Subscript[V, 1]))=((Subscript[N, 2])/(Subscript[N, 1]))

Take the voltages to be the rms voltages then we see that the effect of the transformer is to vary the voltage.  A step-up transformer increases the voltage (Subscript[V, 2])>(Subscript[V, 1]) and a step-down transformer decreases the voltage (Subscript[V, 2])<(Subscript[V, 1]).  If we consider the power in the circuit then the instantaneous power is  𝒫=V I, so it follows that there is an inverse proportionality between the voltage and current.

((Subscript[V, 2])/(Subscript[V, 1]))=((Subscript[N, 2])/(Subscript[N, 1]))=((Subscript[I, 1])/(Subscript[I, 2]))

Transformers are very common.  Many transformers have standard household voltage as its input and DC output at a different voltage.  After the transformer a rectifier circuit is used to convert the AC output of the transformer to DC.