Chapter 33 – Electromagnetic Induction

 

This chapter begins to develop the fundamental connection between electric and magnetic fields.  The central notion is that changes in magnetic flux cause a current in a loop of wire.  Remember the electric flux was used in the chapter on Gauss’s Law,

 

ΦE = ∫EAds,

 

where ds is a differential area with a vector sense perpendicular to the surface.  The magnetic flux is defined analogously with B replacing E,

 

ΦM = ∫BAds.

 

Note that these integrals are not over closed surfaces.  The induced currents we are interested in are produced by changes in ΦM,

 

M/dt = emf (Faraday’s Law)

 

The direction of the induced current “opposes” the change in flux (Lenz’ Law).

 

Much of modern technology depends on the physics of induced currents.  So I encourage you to make the effort to work through this chapter carefully.

 

1.         To begin to see how this works, imagine a metal bar moving from left to right through a magnetic field that points into the page.

 

                        x x x                x x x

                                                                       

                        x x x                x x x

 

                        x x x                x x x

 

 

The charges in the bar feel a magnetic force q v x B.  The force on positive charges is up.  So positive charges migrate to the top of the bar leaving a shortage of negative charges on the bottom.  Or if you insist on thinking about electrons as the current carriers, then negative charges migrate to the bottom leaving unbalanced positive charges on top.  In either case, the charges set up an electric field in the bar that points from top to bottom.  The charges stop moving when the electric force just balances the magnetic force, qE = qvB.  Consequently, the magnitude of the electric field in equilibrium is vB.  If the length of the bar is L, then the emf generated across the bar is just EL, or vBL.

 

For me, this is the most intuitively easy to digest example of an induced emf.  This emf can produce an induced current if we imagine the bar in the above example sliding on a U-shaped.  For concreteness, the U is to the left of the bar in the picture above so that the area inside the loop created by the bar and the U-shaped wire is increaseing.  If the resistance in the loop is R, the induced current I is emf/R = vBL/R.

 

2.         The magnetic field exerts a force on the bar due to current running through the bar.  The current is upward and the magnetic force on the sliding bar is I B L and points in the direction of I x B, to the left.  Therefore I have to exert a force IBL to the right to keep the bar moving at a constant speed.  The power I input is Fv which is IBLv or replacing I with vBL/R, (vBL)2/R.  All of the power I input shows up as I2R, power dissipated by the resistance in the loop.  Replace vBl by IR in the previous equation to show this.

 

3.         This simple example illustrates how it is possible to turn mechanical energy into electrical energy with 100% efficiency.

 

4.         This example could also be understood in terms of Faraday’s Law, emf = M/dt.  In this case, the magnetic flux in the loop is increasing because the area of the loop is increasing.  If the area was A0 at t = 0, the area at time t is just,

 

            A(t) = A0 + vtL.

 

ΦM = BA(t) = B(A0 + vtL) so the induced emf = M/dt = BvL.  The same answer we got above.

 

5.         Lenz’ law gives the direction of the induced emf.  The induced current is in the direction that opposes the change in flux.  In this case the flux in the loop is increasing, so the induced current will produce a B field opposite to the B field through which the bar is moving.  Using our right-hand rule, we see that an upward induced B field is produced by a current going counterclockwise in our loop.  This agrees with our earlier analysis of the magnetic field produced by a current carrying loop.

 

6.         Faraday’s and Lenz’ Laws are much more general than the case we just examined because the change in ΦM could be caused by changing the magnetic field instead of the area of the loop.  Or by moving a loop with constant area from a region where B = 0 to a region where B ≠0.

 

7.         Many applications of induced emfs involve physical systems with many loops of wire since that geometry magnifies the effect.  If a system has N loops,

 

            induced emf = N M/dt.

 

8.         To get a better understanding of how Lenz’ Law is used to find the direction of the induced current, carefully review the six examples in figure 33.23 in the text.

 

9.         In generalizing from our simple example of the bar moving through a constant magnetic field to a situation where the elements of the circuit are stationary but the magnetic field is changing, we have introduced a completely different mechanism for generating an induced emf.  In the first case, the force on the charges was given by q v xB, the magnetic force introduced in chapter 32.  But now, for reasons which are not obvious, a changing magnetic field produces a force that acts on stationary charges!  The only force we know about that exerts forces on stationary charges is the electric force, q E.  Consequently it appears that a changing magnetic field produces an electric field.  This conclusion is necessary because of the experimental observation that moving a magnetic into or out of a loop of wire produces an induced emf!

 

10.       In the book, these new electric fields are called “non-Coulomb electric fields” because they are not produced by charges but instead from Faraday’s Law,         N M/dt.

 

11.       Inside of a solenoid B = μ0 I n, where n = N/l=loops/length (l is a lowercase L!).  Note that a solenoid in a circuit is called an “inductor.”  If the current is changing, dB/dt = μ0 n dI/dt.  For definiteness, imagine the current increasing so dI/dt > 0.  The magnitude of the induced emf = N M/dt = NA μ0 n dI/dt, where A is the cross-sectional area of the solenoid.  This emf is in a direction to oppose the increase in magnetic flux caused by dI/dt.  Since I is increasing, the induced emf wants to produce a current in the opposite direction, hence the emf induced across an inductor is emfL = Δ VL = - NAμ0 n dI/dt.  The inductance L is defined so that,

 

Δ VL = - L dI/dt,

 

which leads to L =  NAμ0 n= N20/l for a solenoid.  Just like capacitance, inductance depends only on the geometry of the inductor.

 

12.       Δ VL is the back emf produced by an inductor on itself.  If instead, we imagine a second loop wrapped around the solenoid, a back emf will be produced in that second loop in response to the changing magnetic flux through the solenoid.  The back emf on the second loop, Δ V2 = - N20 n1 dI/dt where n1 = N1/L and it is assumed that both coils are wrapped around the same cylinder so A is the same as the previous case.  The ratios of the two back emfs, ΔV2 /ΔVL = N2/N1.  This situation basically defines a transformer where an emf can be increased or decreased depending on the ratio of N’s.  In a real transformer, the solenoid has an iron core because iron, being magnetic, increases the magnetic field inside the solenoid making the flux much larger which increases the induced emfs.

 

13.       The work done pushing a charge through an inductor against the induced emf is,

 

Work done by Battery = - dq ΔVL,  where  ΔVL = -LdI/dt. 

 

Since I = dq/dt, the small amount of charge dq = I dt.  The work done moving that small charge through the inductor = I dt L dI/dt = I L dI.  To find the total work done by the battery as the current goes from 0 to I, integrate IL dI from 0 to I to get ˝ LI2.  The energy stored in an inductor is UL = ˝ LI2.  For a solenoid, L = N20/l and I = B/μ0n.  Putting those back into UL you get, UL = (B2/2μ0) Al, Al equals the volume inside the solenoid.  So the energy density in the solenoid = uL = B2/2μ0 which is eerily similar to the energy density inside a capacitor = uC = ε0E2/2.

 

14.       For a circuit with an inductor and capacitor in series, the current through the inductor and the charge on the capacitor oscillate with frequency = ωL = [1/LC]1/2.  When the current is a maximum, the charge is a minimum and when the current is a minimum the charge is a maximum.  Assuming the circuit begins with the capacitor fully charged, q(t) = qmax cos(ωL t).  The current is -dq/dt = i(t),

 

i(t) =  ωL qmax sin(ωL t), where the maximum current is ωL qmax .

 

15.       For a circuit with an inductor and a resistor, the current as a function of time is,

 

i(t) = i(0)exp(-Rt/L), where i(0) is the current at the start of problem.