Download Week 10 - Electromagnetic Induction

Week 10 - Electromagnetic Induction
October 28, 2012
Exercise 10.1: Discussion Questions
a) Two circular loops lie side by side in the same plane. One is connected to a source that supplies an
increasing current; the other is a simple closed ring. Is the induced current in the ring in the same
direction as the current in the loop is connected to the source, or opposite? What if the current in
the first loop is decreasing? Explain.
Answer:
The best ting here is to draw a sketch of the situation. Let’s for the sake of the argument suppose
that the current is going clockwise in the first coil. As the current is increasing in the first coil, the
magnetic field is increasing trough the second coil from below. By Lenz’ law the current will therefore
flow in a direction which opposes this increasing magnetic field. If it has to oppose a magnetic field
coming from below it has to be going in the same direction as the current in the first coil.
But if the current in the first coil is decreasing the magnetic field coming from below trough the
second coil is decreasing in strength. Lenz’ law tells us the current will again flow to oppose this
decrease in strength and thus the current in the second coil will now have to flow the opposite way
of the first coil.
b) A sheet of copper is placed between the poles of an electromagnet with the magnetic field perpendicular
to the sheet. When the sheet is pulled out, a considerable force is required, and the force required
increases with speed. Explain.
Answer:
As the copper sheet is placed between the poles of the electromagnet there is a considerable magnetic
flux trough it. When this sheet is now pulled out of the field, this flux decreases and by Lenz’ law
there will be currents in the sheet that flows in such a way that opposes the decrease in flux. These
current will now feel a magnetic force proportional to the speed at which you pull the sheets out of
the magnetic field. This is the reason for the force felt.
Draw the situation to ensure yourself that this force is indeed in the direction opposite to the motion
of the sheet.
Note that the same phenomenon would not happen if you had a material with a very high resistance.
Current would then not be able to flow and there would not be a corresponding magnetic force.
c) A student asserted that if a permanent magnet is dropped down a vertical copper pipe, it eventually
reaches a terminal velocity even if there is no air resistance. Why should this be? Or should it?
Answer:
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One explanation comes about when viewing the pipe as a series of vertically stacked circular coils of
copper wire. When the permanent magnet falls toward one such coil it’s magnetic flux is increasing,
so it sets up an eddy current creating a magnetic field which opposes that of the permanent magnet.
The resulting magnetic forces on the permanent magnet from this field is indeed one which opposes
it’s motion. And this field will be larger the faster the magnet falls which indicates that there comes
a velocity for which the force is equal to gravitational force.
Exercise 10.2: Search Coils and Credit Cards
A search coil consists of N loops, each with area A, connected trough a circuit of resistance R to an
ammeter. The search coil is first placed such that the area vector A is parallel with the magnetic field
lines in a certain region. Then it is turned in a short time ∆t such that A is now perpendicular to the
magnetic field lines and there is no flux trough the search coil.
a) Derive the equation relating the total charge Q flowing trough the search coil during the time ∆t to
the magnetic field magnitude B.
Solution:
When the area vector is parallel with the field the flux trough the coil is φB = BN A and when it is
perpendicular to the field the flux is zero. The average rate of change of the flux is then related to
the average electromotive force trough Faraday’s law
BN A
∆φB
=
= ε = IR
∆t
∆t
(1)
now the average current is equal to the total charge divided by the time interval the current flows,
I = Q/∆t, so that when we substitute this into the above equation and rearrange we get
Q=
N BA
.
R
(2)
Q=
N BA
.
R
(3)
Answer:
b) In a credit card reader, the magnetic strip on the back of the credit card is rapidly ’swiped’ past a
coil within the reader. Explain, using the same ideas that underlie the operation a search coil, how
the reader can decode the information stored in the pattern of magnetization on the strip.
Answer:
When the different regions of different magnetization swipe past this coil they induce a current in
it. This current is dependent on the magnetic field strength from different parts of the magnetized
pattern, but it should also be dependent on how fast you run your card trough the reader.
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c) Is it necessary that the credit card be ’swiped’ trough the reader at exactly the right speed? Why or
why not?
Answer:
From the answer it (b) it seems like it should depend on exactly that, since the current is proportional
to the velocity of the card. However it might be the ’shape’ of the signal produced that matters.
Alternatively, from the relation in (a), the total charge is independent of the time period of the swipe.
So card readers can read the total charge and give an identification from that.
Exercise 10.3: Make a Generator?
You are shipwrecked on a deserted tropical island. You have some electrical devices that you could
operate using a generator, but you have no magnets. The earth’s magnetic field at your location is
horizontal and has magnitude 8.0 × 10−5 T, and you decide to try to use this field for a generator by
rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 V and
estimate that you can rotate the coil at 30 rpm by turning a crank handle. You also decide that to have
an acceptable coil resistance, the maximum number of turns the coil can have is 2000.
a) What area must the coil have?
Solution:
Our estimate for 30 rpm implies an angular frequency of
ω = 30 ×
2π
= π s−1 .
60 s
(4)
Then for any angle the flux trough the coils is
φB = BN A cos ωt
(5)
dφB
= πBN A sin ωt = ε
dt
(6)
such that by Faraday’s law
Now the peak voltage εp is obtained when t = 0 so that we obtain
A=
9.0
εp
=
m2 = 17.9 m2 .
N πB
π × 2 × 103 × 8 × 10−5
(7)
A=
εp
9.0
=
m2 = 17.9 m2 .
N πB
π × 2 × 103 × 8 × 10−5
(8)
Answer:
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b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates?
Do you think this device is feasible? Explain.
Solution: Assuming the coil is circular, it has an area πr2 , so we get a radius of
r
r=
17.9
m = 2.4 m.
π
(9)
The speed of a point on the coil is then
v = ωr = π × 2.4 m/s = 7.5 m/s.
(10)
This implies a diameter of almost 5 m which seems very unpractical when rotating it with a crank
handle, but depending on how much and how solid equipment we have on our desert island, it does
not seem impossible. Some magnets would however make our lives much easier.
Answer:
v = ωr = π × 2.4 m/s = 7.5 m/s.
(11)
Exercise 10.4: emf in a Bullet
At the equator, the earth’s magnetic field has a value of 8 × 10−5 T and is approximately horizontal,
pointing toward the north. A solider fighting a war in the region recalls his electromagnetism and starts
to wonder weather it is created an emf in his bullets while in flight. The bullets have a length of l = 1.0 cm
and a diameter of d = 0.4 cm and have a speed of v = 300 m/s while airborne.
a) If the solider shots toward the east, is there an emf? If so what is it’s value and which side (top,
bottom, front or back) is at the higher potential?
Solution:
If the solider is shooting towards the east, the magnetic field will be perpendicular to the velocity of
the bullet. Furthermore the magnetic force given by F = qv × B will be pointing upward (on positive
charges, downward on negative). This magnetic force will force the electrons in the bullet down until
it accumulates a large enough charge on the downside on the bullet so that resulting electric field
exactly balances this magnetic force. In that situation we have that the force on an electron in the
bullet
eE = evB ⇒ E = vB.
(12)
V = Ed = vBd
(13)
This implies a potential difference of
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where d is the diameter of the bullet. The top of the bullet is at the higher potential since all the
positive charge is accumulated there. Numerically this corresponds to an emf of
V = vBd = 3.0 × 102 × 8.0 × 10−5 × 0.4 × 10−4 V = 96 µV.
(14)
V = vBd = 3.0 × 102 × 8.0 × 10−5 × 0.4 × 10−4 V = 96 µV.
(15)
Answer:
b) What is the emf the solider shots toward the south?
Answer: If the bullet is shot toward the south, there will be no magnetic force on the bullet. This
is because the velocity and the magnetic field is anti-parallel.
c) What is the emf between the front and the back of the bullet for any horizontal velocity?
Answer: For any horizontal velocity, the magnetic force will always be perpendicular to it and the
magnetic field. Since both the magnetic field and the velocity lies in the horizontal plane the force
must be pointing up or down. Thus there can not be an emf between the two sides of the bullet.
Figure 1
Exercise 10.5: Terminal Speed
A conducting rod with length L, mass m and resistance R moves without friction on metal rails as shown
in figure 1. A uniform magnetic field B is directed into the plane of the figure. The rod starts from
rest and is acted on my a constant force F directed to the right. The rails are infinitely long and have
negigble resistance.
a) Sketch the speed of the rod as a function of time. Here you might solve the equation of motion for
the speed or use your intuition.
Answer:
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Figure 2
Solution:
As the rod is acted on by the constant force F to the right it gains a velocity in the magnetic field
pointing toward the right. This velocity sets up a motional emf in the circuit. Theis gives rise to a
voltage in the circuit which forces a current to flow. The relation between them are
V = vBL = IR
(16)
so that the current in the circuit and in the rod is at any time given by
I=
BL
v.
R
(17)
This current is going counterclockwise in the circuit since the upper end of the circuit is at the higher
voltage. Now a current trough a segment of length L in a magnetic field feels a force ILB which
in this case is directed to the left! Therefore the net force at the rod at any point in time in the
horizontal direction is
Fnet = F − IBL = F −
(BL)2
v
R
(18)
where F is the constant force applied to the right. This net force is precisely analogous to an object
falling under the gravitational force with air resitance and gives rise to a speed of the kind shown in
figure 2.
What happens here is that as the speed increases, the emf and therfore the current increases, but the
opposing force also increases proportional to the current, and therefore there comes a point for which
the two forces exactly cancel eachother and there is no further acceleration.
b) Find an expression for the terminal speed (the speed when the acceleration of the rod is zero). Explain
why this speed is prorportional to the resistance R?
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Solution:
From (a) the net force is zero exactly when
F−
(BL)2
FR
.
v=0⇒v=
R
(BL)2
(19)
The speed is proportional to R because the greater the resistance is the higher the velocity you need
to cause a large enough motional emf to get a current sufficent for the backward force to balance the
forward constant force.
Answer:
FR
(BL)2
v=
(20)
Exercise 10.6: A Persistent Current
(There is currently no answers or solutions included for this exercise.)
A circular wire loop of radius a and resistance R initially has a magnetic flux through it due to an
external magnetic field. The external field then decreases to zero. A current is induced in the loop while
the exernal field is changing; however, this current does not stop at the instant that the external field
stops changing. The reason is that the current itself generates a magnetic field, whic gives rise to a flux
trough the loop. If the current changes, the flux trough the loop changes as well, and an induced emf
appears in the loop to oppose the change.
a) The magnetic field at the center of the loop of radius a is given by B = µ0 i/2a. If we use the crude
approximation that the field has this same value at all points within the loop, what is the flux of this
field trough the loop?
b) By using Faraday’s law, and the relationship ε = iR, show that ater the external field has stopped
changing, the current in the loop obes the differential equation
di
=−
dt
2R
πµ0 a
i.
(21)
c) If the current has the value i0 at t = 0, the instant that the external field stops changing, solve the
equation in part (b) to find i as a function of time for t > 0.
d) If the loop has radius a = 50 cm and resistance R = 0.10 Ω, how long after the external field stops
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changing will the current be equal to 100
i0 ?
e) In the examples in the book, this effect was ignored. Explain why this is a good approximation.
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