MARCH 11, 2014
• REVIEW OF
ELECTROMAGNETISM
Reminders about the quiz
• Remember to print the approved AP equation pages to take to the
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quiz.
Note that Lenz’ Law and the various RHRs aren’t in the equation
list. You need to be able to apply these from memory. It’s
acceptable to use hand motions for the RHRs during the quiz.
Remember to take a hand calculator.
If you’re using a laptop, have your AC adapter handy.
While taking the quiz, the only application you may have open is
WebAssign.
Quiz Advice
• Review your work, the keys, and my comments for Ch. 22,23
assignments.
• For additional practice beyond what’s provided in this presentation,
see the Chapter Reviews. The practice problems will refresh your
memory in using fundamental relationships such as those for the
magnetic forces on charged particles and currents and for induced
emfs due to changing magnetic flux. The problems will also give
you practice in applying RHRs and Lenz’ Law.
More specific advice
•Review the concept of torque and its application to motors. Torque is
the product of a force and its moment arm. The latter is the
perpendicular distance from the axis of rotation to the line of action of
the force. (See WA E.11.01 and V17.)
•Review the function and operation of the electromagnetic devices
we’ve studied: motors, generators, transformers. Know what
fundamental physical principles they make use of and how.
•When determining the magnetic force on a moving charge in a
magnetic field, start with the general relationship for the magnitude of
the force: Fmag = |q|vBsinq. Note that all symbols in this equation are
magnitudes. Use a RHR to determine the direction of the force.
•In applying Lenz’ Law, remember that this has to do with how a
system responds to a change in magnetic flux. In order to have an
induced emf, the magnetic flux must change. There are 3 possible
changes: change in field strength, change in area covered by the
magnetic field, change in angle between the field lines and the
normal to the area.
Even more specific
•WA L15AL, Item 1. This question dealt with the change in current as
the magnet was being pushed into the coil. The observation was that
as the magnet was pushed in, the current first increased and then
decreased to 0. The current was always >= 0 while the magnet was
being pushed in. The goal of the problem was to explain the
observation. Here is an excellent explanation.
“The induced current was proportional to the rate of change of the
magnetic flux. The main factor which affected flux in this experiment was
the position of the magnet near the coil. The rate of change of the
position, which is velocity, was proportional to the rate of change of flux,
which relates to the induced current. This shows that the current varied
with the velocity, so while the magnet was accelerated into the coil, the
induced current rose with the velocity. When the magnet decelerated in
the coil, the velocity and current both lowered. When the velocity was
once again zero, the current was zero as well.”
-B. Dalla Rosa
Follow up on L14 (field of a bar magnet)
Now consider the problem in L14 about
determining how the net magnetic field
changes along the perpendicular bisector of
the magnet. To the right is shown two
magnetic field lines. First, note how they’re
skewed to the right. This is to be expected
from the fact that the magnet is oriented
perpendicular to the Earth’s magnetic field.
At points P and Q, the magnetic field vectors of the earth, Be, and the bar
magnet, Bbar, are drawn together with the net magnetic field, Bnet. Note that Be
has the same direction and magnitude at both points. This is because there’s
negligible change in the Earth’s field over such a small distance (compared to
the size of the Earth). On the other hand, Bbar decreases significantly with
increasing distance from the magnet. As a result, Bnet not only decreases in
magnitude but also gets steeper at increasing distance. One can imagine that
at very large distances from the magnet, Bbar would be insignificant, and Be =
Bnet.
Follow up on L14, con’t.
By the way, the decrease of magnetic field with distance is probably
the primary reason why the value of the field that you calculated for
the bar magnet was less than the typical value quoted in the text.
The way the field of a bar magnet would typically be measured is
close to a pole. This field is expected to be strongest here. You can
see for yourself that the field lines are dense close to a pole.
Of course, another reason for the difference may simply be that
your magnet is different than the one the text calls typical.
Follow up on L27 (Genecon)
•In this lab, the primary observation is that the Genecon becomes
harder to turn physically the less electrical resistance there is
between the contacts. Here’s an explanation:
If you turn at an approximately constant rate, the rate of change of
magnetic flux and hence, the induced emf, will be approximately
constant. (Faraday’s Law) Since I = V/R with V representing the
induced emf, the current will be inversely proportional to the
electrical resistance. Now consider the mechanical resistance.
With greater current, there will be greater magnetic force on the
windings of the armature. (Fmag = BILsinq) This results in a counter
torque, which opposes the motion of the armature. Thus, the
mechanical resistance to turning depends directly on the current,
which is, in turn, inversely proportional to the electrical resistance.
Solutions for P207
Problem 1. Considering just the initial acceleration of the electrons
between
the two vertical plates, determine an equation for v0 in terms of e,
Low
m, and
V1, where m is the mass of an electron.
voltage
here
a, v
coils
P207, Problem 1, con’t.
Strategy: This can be done either as a net force problem (combined with
a dvat to determined an expression for the acceleration) or a
conservation
of energy problem. We choose the latter below.
Low
voltage
System
– electron and vertical plates
here
Initial state – electron with 0 velocity at left plate
a, v v at right plate
Final state – electron with speed
0
Ext forces – none (gravity ignored)
Wext = DEsys
0 = DK + DUel
= (1/2)m(vf2 – vi2) + qDV
= (1/2)m(v02 – 0) + q(Vf – Vi)
= (1/2)mv02 + q(V1 – 0)
= (1/2)mv02 + (-e)V1 (note the substitution of –e for q)
P207, con’t.
Solving for v0: v0 = (2eV1/m)1/2
Problem
Low2. In order for electrons to move from left-to-right at constant
velocity
between the horizontal plates, what must the direction of the
voltage
magnetic
here field be? Explain your answer.
a, v session that the electric force was up.
We established in the last WebEx
Therefore, the magnetic force must be down to achieve equilibrium.
Using the left hand rule with thumb pointing down and the fingers
pointing in the direction of the velocity, the fingers curl into the screen.
This is the magnetic field direction.
Problem 3. Assuming that the electrons move with constant velocity v0
within the region of electric and magnetic fields in the evacuated tube,
determine an equation for e/m in terms of B, V1, V2, and d only.
This is a net force problem. See Problem 7 of M10c for the solution
leading to v0 = E/B. Substituting E = V2/d and the result of Problem 1 for
v0, we obtain e/m = V22/(2B2d2V1)
P207, con’t.
Problem 4. Now assume that the electric field of the horizontal plates is
turned off, leaving only the magnetic field. The electrons then move in a
circular
arc of radius R. Determine an equation for e/m in terms of V1, B,
Low
andvoltage
R only.
here
This is a net force problem with only a magnetic force. See Problem 7 of
v /(eB). Solving for e/m and substituting
M10b leading to the result R a,
= mv
0
the expression for v0 from Problem 1, the result is e/m = 2V1/(BR)2.
More problems for practice
Next is a series of problems from past AP exams. We’ll use these to review
conceptsLow
and relationships in electromagnetism.
voltage
here
Low
voltage
here
a, v
Low
voltage
here
a, v
This is a conservation of energy problem. With the system
being the cart and the Earth, the only external force being
the normal force of the plane, which does no work, the initial
state being the cart at the highest point and the final state
just before the cart enters the field, you can show that v =
(2gy0)1/2.
Low
voltage
here
a, v
Low
voltage
here
a, v
i.
ii.
The magnitude of the induced emf = Df/Dt = D(BAcosq)/Dt. The
normal to the side of the cart is parallel to the field, hence, q = 0o. B
is constant. Thus the emf becomes BDA/Dt. The area is A = hw, and
h is constant in the field. Thus, induced emf = Bh(Dw/Dt) = Bhv =
Bh(2gy0)1/2.
The current is I = (induced emf)/R = Bh(2gy0)1/2/R.
Low
voltage
here
a, v
Low
voltage
here
a, v
Bhw
Bhw/R
Low
voltage
here
a, v
Low
voltage
here
a, v
Using the RHR where you grasp the wire with your right hand and
your thumb pointing in the direction of the current, your fingers trace
the magnetic field lines. For I1, the field is into the screen at point P.
For I2, the field is out of the screen. The field due to I1 is stronger since
r1 < r2. Therefore, the net field is in the same direction at point P as the
field of I1, into the screen.
Low
voltage
here
a, v
Low
voltage
here
a, v
Bnet = B1 – B2, where the positive direction is into the screen.
Substituting into the given formula,
Bnet = (m0/2p)(I1/r1 – I2/r2)
Low
voltage
here
a, v
Low
voltage
here
a, v
The force that the left wire exerts on the right wire is F12 = B1I2Lsinq.
Note that the field used is that of the wire exerting the force, while the
current used is that of wire that the force is being exerted on. The angle
between the current and the field is always 90o; therefore, F12/L = B1I2.
Check: The force per unit length that the right wire exerts on the left
would be F21/L = B2I1 by similar reasoning. By Newton’s 3rd Law, F12 =
F21. Therefore B1I2 = B2I1. This equality can be shown to be true using
the equation for the magnetic field of a long, straight wire: B = m0I/(2pr).
Since r is the same for both wires, B/I is a constant.
Low
voltage
here
a, v
Fmag
Low
voltage
here
a, v
x
Consider when the particle is in the magnetic field only. Using the RHR at the
point of entry into this region, the magnetic force is down (to the center of the
circle) so the thumb points down. Fingers curl into the page in the direction of
the field into the screen. The extended fingers point to the left. Since this is
opposite the direction of motion, the particles must be negative.
Low
voltage
here
a, v
E
Low
voltage
here
a, v
The magnetic force is down between the plates. Therefore the
electric force must be up in order to have equilibrium. Since the
particles are negative, the electric field is opposite the direction of
the electric force.
Low
voltage
here
a, v
Low
voltage
here
a, v
Fel
Fnet = Fel – Fmag
= |q|E - |q|vBsinq
0 = E – vB
= V/d – vB
V = vBd = (1.9E6 m/s)(0.20 T)(6.0E-3 m)
= 2300 V
+y
Fmag
Low
voltage
here
a, v
Low
voltage
here
Fmag
a, v
Take + to the left in the direction of the magnetic force shown on the
diagram. Do a net force analysis to determine q/m.
Fnet = Fmag
mv2/r = |q|vBsinq
|q|/m = v/(rB)
= (1.9E6 m/s)/[(0.10 m)(0.20 T)
= 9.5E7 C/kg