29a. Electromagnetic Induction
Conceptual Induction
Part A
What is the direction of the induced current in the loop when the loop is above the
solenoid, moving downward?
clockwise
Correct
Part B
What is the direction of the induced current when the loop is at the midpoint of the
solenoid and still moving downward?
no currentCorrect
Part C
What is the direction of the induced current when the loop is below the solenoid and
moving downward?
counterclockwise
Correct
Introduction to Faraday's Law
Part A
Consider the direction of the electric field in the figure. Assume that the magnetic field
points upward, as shown. Under what circumstances is the direction of the electric field
shown in the figure correct?
if
decreases with time
Correct
Part B
Now consider the magnetic flux through a surface bounded by the loop. Which of the
following statements about this surface must be true if you want to use Faraday's law to
relate the magnetic flux to the line integral of the electric field around the loop?
The surface can be any surface whose edge is the loop.
Correct
You are free to take any surface bounded by the loop as the surface over which to
evaluate the integral. The result will always be the same, owing to the continuity of
magnetic field lines (they never start or end anywhere, since there are no magnetic
charges).
It is important to understand the vast differences between electric fields produced by
changing magnetic fields via Faraday's law and the more familiar electric fields produced
by charges via Coulomb's law. Here are some short questions that illustrate these
differences.
Part C
When can an electric field be measured at any point from the force on a stationary test
charge at that point?
no matter how the field is generatedCorrect
In fact, this operation defines an electric field. Similarly, if the test charge is moving, it
will measure magnetic fields.
Part D
When can an electric field that does not vary in time arise?
in either of the above two cases
Correct
Part E
When will the integral
around any closed loop of the projection of the electric
field along that loop be zero?
only if the field is generated by the coulomb field of charges that are static or moving in a
straight line
Correct
The electric field generated by a uniformly moving charge is just the moving field of a
static charge, which always has zero loop integral.
Here is a simple quantitative problem that uses Faraday's law.
Part F
A cylindrical iron rod of infinite length with cross-sectional area is oriented with its
axis of symmetry coincident with the z axis of a cylindrical coordinate system as shown
in the figure. It has a magnetic field inside that varies according to
the theta component
of the electric field at distance
larger than the radius of the rod.
Express your answer in terms of
and .
,
,
,
. Find
from the z axis, where
, and any needed constants such as
is
, ,
=(-A*B_1)/(2*pi*R)Correct
29b. Electromagnetic Induction
Rail Gun
Part A
A conducting rod is free to slide on two parallel rails with negligible friction. At the right
end of the rails, a voltage source of strength in series with a resistor of resistance
makes a closed circuit together with the rails and the rod. The rails and the rod are taken
to be perfect conductors. The rails extend to infinity on the left. The arrangement is
shown in the figure.
There is a uniform magnetic field of magnitude , pervading all space, perpendicular to
the plane of rod and rails. The rod is released from rest, and it is observed that it
accelerates to the left. In what direction does the magnetic field point?
out of the plane of the figureCorrect
Part B
Assuming that the rails have no resistance, what is the most accurate qualitative
description of the motion of the rod?
The rod will accelerate but the magnitude of the acceleration will decrease with time; the
velocity of the rod will approach but never exceed a certain terminal velocity.
Correct
Part C
What is the acceleration
of the rod? Take
Express your answer as a function of
mass of the rod .
,
to be the mass of the rod.
, the velocity of the rod
, ,
, and the
=(((V-v_r(t)*B*L)/R)*B*L)/mCorrect
Making the substitution
, you obtain the dfferential equation
,
which you can solve to find the velocity of the rod as a function of time:
.
To achieve a high acceleration, which is necessary for a useful gun, a magnetic field of
large magnitude and a high voltage are advantageous.
Part D
What is the terminal velocity reached by the rod?
=V/(B*L)Correct
A larger magnetic field increases the acceleration of the rod, but lowers the terminal
velocity: a trade-off for rail gun engineers!
The Ampère-Maxwell Law
Part A
First find
, the line integral of around a loop of radius located just outside
the left capacitor plate. This can be found from the usual current due to moving charge in
Ampère's law, that is, without the displacement current.
Find an expression for this integral involving the current
constants such as .
and any needed physical
=mu_0*I(t)Correct
Part B
Now find an expression for
, the same line integral of
around the same loop of
radius located just outside the left capacitor plate as before. Use the surface that passes
between the plates of the capacitor, where there is no conduction current. This should be
found by evaluating the amount of displacement current in the Ampère-Maxwell law
above.
Express your answer in terms of the electric field between the plates
plate area
,
, the
, and various physical constants.
=mu_0*epsilon_0*dE(t)/dt*ACorrect
A necessary consistency check
Part C
We now have two quite different expressions for the line integral of the magnetic field
around the same loop. The point here is to see that they both are intimately related to the
charge
terms of
on the left capacitor plate. First find the displacement current
.
Express your answer in terms of
=dq(t)/dtCorrect
,
, and other physical constants.
in
Part D
Now express the normal current
.
Express your answer in terms of
in terms of the charge on the capacitor plate
,
, and any needed physical constants.
=dq(t)/dtCorrect
Using Gauss's law, you have shown that the displacement current from the changing
electric field between the plates equals the current from the flow of charge through the
wire onto that plate. This means that the Ampère-Maxwell law can consistently treat
cases in which the normal current due to the flow of charge is not continuous. This
realization was a great boost to Maxwell's confidence in the physical validity of his new
displacement-current term.