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,
dΦM/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
= dΦM/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 = dΦM/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 dΦM/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 dΦM/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 dΦM/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= N2
Aμ0/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 = - N2Aμ0 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
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 = N2 Aμ0/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.