Download Electromagnetic Induction

Electromagnetic Induction (approx. 1.5 h) (11/9/15)
Introduction
In 1819, during a lecture demonstration, the Danish scientist Hans Christian Oersted noticed that
the needle of a compass was deflected when placed near a current-carrying wire showing that electric
current created a magnetic field. This led investigators to wonder if a magnetic field could be used to
create electricity. That this is indeed possible was first demonstrated in 1831 by both Joseph Henry in
America and Michael Faraday in England. The phenomenon is known as electromagnetic induction and
its mathematical statement is commonly referred to as Faraday’s law of induction. It is employed in
electric power generation, voltage transformers, electronic circuits and many other technologies.
In this lab you will investigate how magnetic fields can be used to induce electrical currents in a
solenoid (cylindrical coil of wire). You will make qualitative measurements and then determine the
extent to which your observations confirm Faraday’s law and Lenz' law which are described in the
theory section at the end of the lab.
Equipment
 Agilent (HP)
power supply
 galvanometer
 10kΩ resistor



copper primary and
secondary coils
alligator clips
multimeter




2 banana-to-alligator leads
D battery (in holder)
air core solenoid
steel & Al(or brass) rods



compass
spring switch
bar magnets
(1 reg,1 weak)
SAFETY NOTE
Do not connect or disconnect wires unless the power supply is off.
For Your Report: Include a cover page and introduction. Label sections clearly, corresponding to
headings in this handout. You should take the time during each part of the lab to record your
observations and conclusions (after discussion with your lab partners) and write them out using
complete sentences before proceeding to the next section. Write an overall summary for your
conclusions and state whether your observations support the theory of electromagnetic induction.
Procedure
A. The Direction of the Current and the Direction of Deflection of the Galvanometer Needle
The galvanometer is an instrument for detecting and/or measuring very small currents. A typical
galvanometer works by responding to the torque exerted by a magnetic field on a current-carrying coil.
(Analog ammeters and voltmeters are based on the same principle.) The reading is displayed by means
of the deflection of a pointer over a scale. The pointer is attached to a small coil which is attached to a
spring and placed between the poles of a magnet. Current flowing through the coil causes it to deflect
due to the torque created on it by the magnetic field. The deflection is to the right or left depending on
the direction of the magnetic field of the coil and thus on the direction of the current. In this section you
will establish how the direction of the current is related to the direction of deflection.
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Electromagnetic Induction
I
-
+
10kΩ
D Battery
Since galvanometers have very low resistance and
are designed only for very small currents it is
important not to connect them directly to a voltage
source without making sure the current will be limited
to an acceptable range. Since the galvanometer you will be using has a range of only ±500µA
(microamperes), make sure that the resistor you place in series with your 1.5V battery will limit the
current to be no more than ±500µA. (Measure the resistance of your resistor and use Ohm’s law to
calculate the maximum current!) CAUTION: EXCESSIVE CURRENT WILL DAMAGE THE
GALVANOMETER. IF YOU ARE NOT CERTAIN THAT YOU HAVE CONSTRUCTED
YOUR CIRCUIT CORRECTLY, ASK THE INSTRUCTOR TO CHECK IT BEFORE
CONNECTING TO THE GALVANOMETER. Since you know that the direction of the current is
out of the positive terminal and into the negative terminal (i.e. base) of the battery, you can find how the
direction of deflection corresponds to the direction of current. Reverse the leads to see that the needle
deflects in the opposite direction. Record your observations below and in your report:
Figure 1: Circuit using a battery and a resistor to
determine how galvanometer deflection direction
corresponds to current direction.
Direction of current into galvanometer
needle deflects
right or left
When the current goes into the left terminal of the galvanometer, the needle deflects
When the current goes into the right terminal of the galvanometer, the needle deflects
Note: For the rest of the lab you will need two bar magnets of different
strengths. Estimate the relative strengths of the two bar magnets by feeling
the attractive forces with which they stick to a steel object.
N
S
B. Magnetic Field Around a Bar Magnet
Lay the stronger bar magnet flat on the table and use a compass to
investigate the shape of the magnetic field around the bar magnet. Recall Figure 2: Sketch the
that the compass needle points along the magnetic field with the North end field lines around the
pointing in the direction of the field. The intensity can be estimated by bar magnet
noting how quickly the compass needle responds. In your notes you should
record the direction and relative intensity of the field around the North pole of the magnet, near the
center, and near the South pole. Make a sketch of what you suppose the magnetic field lines look like
around the bar magnet. (Use arrows to indicate direction and closeness of lines to indicate field
strength.)
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Electromagnetic Induction
C. Using Bar Magnets to Induce Current in a Solenoid
You should have one pair of copper primary and secondary coils.
The “primary” coil has about 250 turns and fits inside the
“secondary” coil, which has about 1900 turns. You should also
have an air-core solenoid, which is labeled with the number of turns
it has. (Some air-core solenoids have both 220 turn and 440 turn
coils. Although they may appear to be single solenoids, they
actually consist of two sets of coils, one inside the other. If you have
G
one of these air-core solenoids you should use the 440 turn
connections for this lab.)
Connect the galvanometer to the terminals of your air-core solenoid. Figure 3: Using a bar
(There is NO POWER SUPPLY and no resistor in this circuit!) magnet to induce current
Move the stronger bar magnet into and out of the coil (Fig. 3), in a solenoid.
noting and recording the effects in the data table on page 6. You
should note the effect of 1) the speed with which you move the magnet and, 2) changing the polarity
of the magnet.
Insert the north pole as quickly as possible. Pause when it is inserted about half way. Note the maximum
galvanometer deflection and the deflection when the magnet is stationary inside the solenoid. Repeat
your motion several times in order to verify your estimate of the maximum deflection.
Start with the North pole inserted halfway, then withdraw it as quickly as possible. Again record the
direction and estimate the maximum deflection.
Repeat, this time moving the magnet more slowly.
Repeat, for both fast and slow motions, this time using the South poles of the magnets.
Repeat using the 1900 turn coil of the primary/secondary coil set and note the differences.
Repeat the above procedures using the weaker magnet and note the differences.
Summarize your results and draw conclusions based on your observations.
D. Investigating the Magnetic Field of a Solenoid
Inside a solenoid the magnetic field is fairly uniform. It can be shown using Ampere’s law that, for a
long solenoid, it is about equal to B  0 nI , where 0  4  10 7 TAm , n is the number of turns of
wire per unit length (of solenoid) and I is the current in the wire. In this section we will investigate
the magnetic field outside of a current-carrying solenoid.
With the power supply off, connect the positive terminal of the DC power supply to one terminal of a
spring switch. Connect the other side of the switch to a terminal of the large, 1900 turn copper
solenoid (i.e. the “secondary” coil). Connect the second terminal on the large copper solenoid to the
negative terminal of the power supply (i.e. as shown in the top part of Fig.4, but using the secondary
coil). Examine the solenoid and determine which way current will flow around the core when power
is supplied.
Lay the large solenoid down on its side. Turn the power supply on and turn the voltage and current
knobs to maximum. Hold a compass near the center of one end of the solenoid. CLOSE the spring
switch (Current flows!). Investigate the magnetic field around the solenoid with the compass,
holding it at about the height of the center of the solenoid. You should be able to determine the
direction of the magnetic field as it goes through the solenoid. How does the direction of the
magnetic field inside the solenoid compare to the direction the current goes around the solenoid?
How does the overall shape of the magnetic field outside a current-carrying solenoid compare to the
shape of the magnetic field outside of a bar magnet (as determined in part B)? (similar or not similar)
Open the switch and reverse the leads on the solenoid. Repeat the previous step and verify that
reversing the current reverses the direction of the magnetic field.
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Electromagnetic Induction
E. Changing Current in One Solenoid to Induce Current in a Second Solenoid
We can create a changing magnetic field inside a solenoid
S
by changing the current. If this solenoid is located
inside a second solenoid, we can induce current in the +
Primary
second solenoid as we change the current, and
Coil
V
therefore the flux, in the first solenoid. (This is the
principle involved in the design of voltage
transformers.)
Disconnect the larger (secondary) coil and replace it with
Secondary
the primary coil as shown in Fig. 4.
Coil
Set the power supply current range button to the 2 A
G
setting.
Connect the large secondary coil to the galvanometer as
shown.
Figure 4: Changing current in the
Place the small copper coil inside the larger copper primary coil will cause a changing flux
solenoid.
in the secondary.
You will be changing the current in the smaller
(“primary”) coil and observing the effect on the current in the larger (“secondary”) coil.
Close the switch, sending current through the primary coil, and observe the effect on the current in the
secondary coil by watching the galvanometer. Note the direction of deflection and estimate the
magnitude of largest deflection. (Close and open the switch several times to make a better estimate.)
Leave the switch closed for a few seconds. What is the deflection of the galvanometer?
Open the switch again, noting the effect on the current in the secondary coil.
Repeat as needed to verify your observations and to allow all lab partners to observe and record the
results.
WITH THE SWITCH OPEN (no current) change the direction of the current in the primary coil by
switching the leads. Again close and open the switch to start and stop the current, noting the effect
on the current in the secondary coil.
Summarize your results and draw conclusions.
F. Effect of Ferromagnetic Material in the Core
When a magnetic field exists in a material, rather than in vacuum, the magnetic field will be changed by
the magnetic properties of the material. Here you will compare the magnetic properties of aluminum (or
brass) and steel. (Note: steel is an alloy composed primarily of iron, a ferromagnetic material.)
Try to pick up both the steel and the aluminum (or brass) rods with your bar magnet. Does it make a
difference whether you use the North or South pole of the bar magnet?
Use a circuit like shown above, but with the air-core solenoid instead of the 1900 turn secondary. WITH
THE SWITCH OPEN, insert the aluminum (or brass) rod inside the 250 turn primary coil. Close and
open the switch and record the maximum current in the secondary (i.e. air-core) coil.
WITH THE SWITCH OPEN, replace the aluminum rod with a steel rod (shiny, with handle). Close and
open the switch and record the maximum current in the secondary coil.
Summarize and draw conclusions about the effects of (ferromagnetic) steel vs. aluminum (or brass).
When a magnetic field exists in a material, rather than in vacuum, the constant  0 is replaced by a value
 that depends on the material. Speculate on the value of  (relative to  0 ) for steel and for
aluminum (or brass).
To observe an interesting effect, try holding the shiny, smooth steel bar (with the “handle”) about
halfway inside the primary. Put the current range setting to 3 A and turn the voltage and current knobs
to maximum. Switch the current on and see what happens.
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Electromagnetic Induction
G. Effect of number of turns of wire in the primary and secondary.
Remove the steel bar and continue to use the smaller copper coil (250 turns) as the primary and the
larger copper coil as the secondary (1900 turns). Close and open the switch and note the effects. With
the switch open, change the connections so that the other air-core solenoid is used as the secondary.
Close and open the switch and note the effects. With the switch open, move the supply connections
from the small copper coil to the leads for the air-core solenoid, so that it becomes the primary, and
connect the small, 250 turn coil to the galvanometer, so it becomes the secondary. Close and open the
switch and record the maximum deflection. Repeat, using the larger, 1900 turn copper coil as the
primary and record your results.
Summarize your results and draw conclusions about the effect of the relative number of turns in the
primary and secondary coils.
H. Questions & Conclusions:
How does the direction of the induced current depend on
which pole of the permanent magnet is inserted into the
solenoid?
How does the magnitude of the induced current depend on the
strength of the bar magnet?
A2
A1
The magnetic field inside a solenoid is given by: B  0 nI .
Outside the solenoid the field is much weaker and may be Figure 5: Two concentric
approximated as zero. Suppose that a solenoid of area A1 solenoids.
and turns per unit length n1 is placed inside a second
solenoid of area A2 , turns per unit length n2 and length L2 (total number of turns in second coil N2=
dI 1
L2n2). Explain why the emf,  , induced in the second coil by a changing current,
, in the first
dt
dI
dI
coil is given by    N 2 A1  0 n1 1 . (If the change is at a constant rate, then 1 may be replaced
dt
dt
I
by 1 )
t
If the secondary solenoid is in a circuit with resistance R, there will be an induced current, I 2   R .
Discuss how the equations predict the direction of the current in the secondary coil depending on
whether the current in the primary coil, I1, is increasing or decreasing.
Does this match your observations? (Hint: Consider the effect of the sign of
dI 1
. Is it increasing or
dt
decreasing?)
Given the expression above, how do the equations predict the induced current depends on the number of
turns in the primary?
In the secondary?
Does this match your observations?
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Electromagnetic Induction
Using Bar Magnets to Induce Current in a Solenoid
Air-Core Solenoid
(# turns________)
Magnet 1 :
(Stronger Bar Magnet)
North Pole
Moves quickly inward
Stationary inside
Moves quickly outward
Moves slowly inward
Moves slowly outward
South Pole
Moves quickly inward
Stationary inside
Moves quickly outward
Moves slowly inward
Moves slowly outward
Magnet 2 :
(Weaker Bar Magnet)
North Pole
Moves quickly inward
Stationary inside
Moves quickly outward
Moves slowly inward
Moves slowly outward
South Pole
1900 Turn Secondary
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Maximum
Deflection, A
Direction of
Current
Moves quickly inward
Stationary inside
Moves quickly outward
Moves slowly inward
Moves slowly outward
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Electromagnetic Induction
Changing Current in One Solenoid to Induce Current in a Second Solenoid
Primary=small copper coil
Secondary=large copper coil
Initial Current Direction
Opposite Current Direction
Maximum
Deflection, A
Maximum
Deflection, A
Direction
of Current.
Direction of
Current.
While closing switch
Current constant
While opening switch
Magnetic Effect of Materials
Primary=small copper coil (250 turns)
Secondary= large copper coil (1900 turns)
Aluminum Rod
Steel Rod
Maximum
Deflection, A
Maximum
Deflection, A
Direction of
Current.
Direction of
Current.
While closing switch
Current constant
While opening switch
Effect of number of turns in primary and secondary
Primary=small copper coil (250 turns)
Secondary = large copper coil (1900 turns) or air-core coil
Secondary=1900 turns
Primary= small copper coil = 250 turns
Maximum
Direction
Maximum
Direction of Current.
Deflection, A of Current Deflection, A
(see above)
(see above)
While closing switch
Current constant
While opening switch
Primary=1900 turns
Secondary= 250 turns
Maximum
Direction of
Deflection, A Current.
While closing switch
Current constant
While opening switch
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Primary = air-core coil = ______turns
Secondary = 250 turns
Maximum
Direction of
Current.
Deflection, A
Electromagnetic Induction
Theory (No Calculus)
The word "flux" comes from the Latin word for "flow" and in
physics refers to the "flow" of a vector quantity through an area.
B
A
A current of water can be described by a vector since it has a

magnitude (amount of water per time) and a direction (direction
of the current). If you image placing a tennis racket in the water
then the "flux" of water through the head of the racket would
depend on 1) the strength of the current, 2) the area of the racket,
and 3) the angle between the area of the racket and the direction
of the current. Electric and magnetic fields are vector quantities
and so we can define either an electric or a magnetic flux: the
amount of electric or magnetic field “passing through” an area. It Figure 6: If the magnetic flux
depends on the strength of the field, the area through which it through any closed loop is
passes and the angle between the area and the field. For a fixed changing as a function of time, an
area, A¸ and a magnetic field given by B, the magnetic field flux induced voltage will be created.
is given by
(fixed, flat area and constant magnetic field)
 B  BA cos 
The angle,  is the angle between the area vector (normal to plane of area) and the magnetic field, both
of which are shown in Figure 6.
Electromagnetic Induction: Faraday’s Law of Induction states that if the magnetic flux through any
closed loop is changing as a function of time, then a voltage or electromagnetic force (emf) will be
induced around that loop.
B
(changes at a constant rate)
   
t
If the loop is part of a closed circuit, then the induced emf will create an induced current. In this lab you
will look at the current induced in a coil of wire (a solenoid) as the magnetic flux inside the solenoid is
changed. If the magnetic field goes down the axis of a solenoid with N coils of wire, then, when the
magnetic field changes at a constant rate, the induced emf is given by
(note: this equation is a special case as described above!)
   NA Bt
The direction of the induced voltage or current is predicted by Lenz' Law which is described below.

Lenz’ Law
In 1834, soon after the discovery of Faraday’s law of induction, the German physicist Heinrich Friedrich
Lenz devised a rule for determining the direction of an induced current in a loop: An induced current
has a direction such that the magnetic field due to the current opposes the change in the magnetic flux
(not the flux itself). Furthermore, the direction of the induced emf associated with the induced current is
that of the induced current. For example, if the flux is increasing, the induced emf will be in a direction
that would cause current to flow around the loop in a direction that would create a magnetic field
opposing the existing magnetic field and thereby decrease the flux. If the flux is decreasing, on the other
hand, the induced current would resist this change by creating a magnetic field which adds to the
existing one.
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Electromagnetic Induction
Theory (Calculus)
Magnetic Flux: In discussing Gauss’ Law we have already defined electric flux. We can similarly
define magnetic flux as the integral of the magnetic field passing through
a surface. When the magnetic

field is uniform over a flat surface
 characterized by the area vector, A , the integral reduces to the scalar
product of the magnetic field, B , and area vector:
   
 B   B  dA  B  A
Electromagnetic Induction: Faraday’s Law of Induction states that if the magnetic flux through any
closed loop is changing as a function of time, then a voltage or electromagnetic force (emf) will be
induced around that loop.
d
   dt B
For the case of uniform magnetic field over a flat area, we can apply the chain rule to show that
d B
dBA 
d  
dB
dA


   A  B   A cos BA
 B cos  BA
 B A sin  BA


dt
dt
dt
dt
dt 
Thus an induced voltage will be observed if:
dB
 the magnetic field is changing as a function of time, so the term A cos  BA
is non-zero
dt
dA
 the area is changing as a function of time, so the term B cos  BA
is non-zero
dt
 the angle between the area vector and field is changing as a function of time, so the term
d
B A sin  BA BA is non-zero.
dt
If the loop is part of a closed circuit, then the induced voltage will create an induced current whose
magnitude depends on the resistance in the circuit. If the flux is through a coil of N turns of wire, then
the flux and resulting emf are simply multiplied by N.
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