Electromagnetic Induction
PHY232
Remco Zegers
zegers@nscl.msu.edu
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
previously:
¾ electric currents generate magnetic field. If a current is flowing
through a wire, one can determine the direction of the field with the
(second) right-hand rule:
¾ and the field strength with the equation: B=μ0I/(2πR)
¾ For a solenoid or a loop (which is a solenoid with one turn):
B=μ0IN/(2R) (at the center of the loop)
If the solenoid is long: B=μ0In (at the center of the solenoid)
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now:
¾ The reverse is true also: a magnetic field can generate an
electrical current
¾ This effect is called induction: In the presence of a
changing magnetic field, and electromotive force
(voltage) is produced.
demo: coil and galvanometer
Apparently, by moving
the magnet closer to the
loop, a current is produced.
If the magnet is held stationary,
there is no current.
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a definition: magnetic flux
¾ A magnetic field with strength B passes through a loop
with area A
¾ The angle between the B-field lines and the normal to the
loop is θ
¾ Then the magnetic flux ΦB is defined as:
Units: Tm2 or Weber (W)
lon-capa uses Wb
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example: magnetic flux
¾ A rectangular-shaped loop is put perpendicular to a
magnetic field with a strength of 1.2 T. The sides of the
loop are 2 cm and 3 cm respectively. What is the
magnetic flux?
¾
B=1.2 T, A=0.02x0.03=6x10-4 m2, θ=0.
¾ ΦB=1.2 x 6x10-4 x 1 = 7.2x10-4 Tm
¾ Is it possible to put this loop such that the magnetic flux
becomes 0?
¾ a) yes
¾ b) no
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Faraday’s law:
¾ By changing the magnetic flux ΔΦB in a time-period Δt a
potential difference V (electromagnetic force ε) is
produced
Warning: the minus sign is never used in calculations. It is
an indicator for Lenz’s law which we will see in a bit.
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changing the magnetic flux
¾ changing the magnetic flux can be done in 3 ways:
¾ change the magnetic field
¾ change the area
¾ changing the angle
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example
¾ a rectangular loop (A=1m2) is moved
into a B-field (B=1 T) perpendicular
to the loop, in a time period of 1 s.
How large is the induced voltage?
x
x
x
x
x
x
x
x
x
x
x
x
• While in the field (not moving) the area is reduced to 0.25m2 in 2 s.
What is the induced voltage?
•This new coil in the same field is rotated by 45o in 2 s.
What is the induced voltage?
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Faraday’s law for multiple loops
¾ If, instead of a single loop, there are multiple loops (N), the
the induced voltage is multiplied by that number:
N
S
demo: loops.
If an induced voltage is put over
a resistor with value R or the
loops have a resistance, a current
I=V/R will flow
resistor R
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lon-capa
¾ You should now try problems 2,3,4 & 7 from
lon-capa set 6.
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first magnitude, now the direction…
¾ So far we haven’t worried about the direction of the
current (or rather, which are the high and low voltage
sides) going through a loop when the flux changes…
N
S
direction of I?
resistor R
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Lenz’s Law
¾ The direction of the voltage is always to oppose the
change in magnetic flux
when a magnet approaches the
loop, with north pointing towards
the loop, a current is induced.
As a results a B-field is made by the
loop (Bcenter=μ0I/(2R)), so that the field
opposes the incoming field made
by the magnet.
demo: magic loops
Use right-hand rule: to make a field
that is pointing up, the current must
go counter clockwise
The loop is trying to push the magnet away
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Lenz’s law II
¾ In the reverse situation where the magnet is pulled away
from the loop, the coil will make a B-field that attracts the
magnet (clockwise). It opposes the removal of the B-field.
Bmagnet Binduced
Bmagnet Binduced
v
v
magnet approaching the
coil
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magnet moving away from
the coil
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left-hand rules
¾ There are several variations of left hand-rules available to
apply Lenz’s law on different systems. If you know them,
feel free to use it. However, they can be confusing and I
will refrain from applying them.
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Be careful
¾ The induced magnetic field is not always pointing
opposite to the field produced by the external magnet.
x
x
x
x
x
x
x
x
x
x
x
x
If the loop is stationary in a field, whose
strength is reducing, it wants to counteract
that reduction by producing a field pointing
into the page as well:
current clockwise
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demo magnet through cooled pipe
S
¾ when the magnet passes
through the tube, a current is
induced such that the B-field
produced by the current loop
opposes the B-field of the
magnet
¾ opposing fields: repulsive force
Binduced
¾ this force opposes the
gravitational force and slow
down the magnet
¾ cooling: resistance lower
current higher, B-field higher,
opposing force stronger
N
vmagnet
S
N
I
Bmagnet
can be used to generate electric energy (and store it e.g. in a capacitor):
demo: torch light
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question
x
A
x
x
x
x
x
x
x
x
x
B
x
x
A rectangular loop moves in, and then out, of a constant magnet field
pointing perpendicular (into the screen) to the loop.
Upon entering the field (A), a …. current will go through the loop.
a) clockwise
b) counter clockwise
When entering the field, the loop feels a magnetic force to the …
a) left
b) right
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lon-capa
¾ you should now try question 5 of lon-capa 6 (you just did
half of that problem).
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Eddy current+demo
¾ Magnetic damping occurs when a flat strip of
conducting material pivots in/out of a
magnetic field
¾ current loops run to counteract the B-field
¾ At the bottom of the plate, a force is directed
the opposes the direction of motion
I
I
x
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x
v
x
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strong opposing force
x
x
x
weak opposing force
x
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x
B-field into the page
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x
x
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v
no opposing force
x
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x
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applications of eddy currents
¾ brakes: apply magnets to a brake disk. The induced
current will produce a force counteracting the motion
¾ metal detectors: The induced current in metals produces
a field that is detected.
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A moving bar
R
B-field into the page
x
x
x
x
x
x
x
x
x
x
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V
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d x
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x
¾ Two metal rods (green) placed parallel at a distance d are
connected via a resistor R. A blue metal bar is placed over the rods,
as shown in the figure and is then pulled to the right with a velocity v.
¾ a) what is the induced voltage?
¾ b) in what direction does the current flow? And how large is it?
¾ c) what is the induced force (magnitude and direction) on the bar?
What can we say about the force that is used to pull the blue bar?
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lon-capa
¾ Now do problems 1 and 6 from lon-capa 6.
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Doing work
¾ Since induction can cause a force on an object to
counter a change in the field, this force can be used to do
work.
¾ Example jumping rings: demo
current cannot flow
current can flow
The induced current in the ring produces a B-field opposite from the one
produced by the coil: the opposing poles repel and the ring shoots in the air
application: magnetic propulsion, for example a train.
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generating current.
¾ The reverse is also true: we can do work and generate
currents
By rotating a loop in a field (by hand, wind
water, steam…) the flux is constantly
changing (because of the changing
angle and a voltage is produced.
θ=ωt with
ω: angular velocity
ω=2πf = 2π/T
f: rotational frequency
T: period of oscillation
NBAωsin(ωt)
demo: hand generator
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Time varying voltage
NBAωsin(ωt)
Vmax
time (s)
C
-Vmax
A
B
B
C
A
side view of loop
¾ Maximum voltage: V=NBA
¾ This happens when the change in flux is largest, which is
when the loop is just parallel to the field
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question
¾ A current is generated by a hand-generator. If the person
turning the generator increased the speed of turning:
¾ a) the electrical energy produced by the system remains
the same
¾ b) the electrical energy produced by the generator
increases
¾ c) the electrical energy produced by the generator
decreased
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Self inductance
L
I
V
¾ Before the switch is closed: I=0, and the magnetic field
inside the coil is zero as well. Hence, there is no magnetic
flux present in the coil
¾ After the switch is closed, I is not zero, so a magnetic field
is created in the coil, and thus a flux.
¾ Therefore, the flux changed from 0 to some value, and a
voltage is induced in the coil that opposed the increase of
current
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L
Self inductance II
I
¾ The self-induced current is proportional to the change in flux
¾ The flux ΦB is proportional to B. e.g. Bcenter=μ0In for a solenoid
¾ B is proportional to the current through the coil.
¾ So, the self induced emf (voltage) is proportional to change in current
L inductance : proportionality constant
Units: V/(A/s)=Vs/A
usually called Henrys (H)
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induction of a solenoid
¾ flux of a coil:
¾ Change of flux with time:
¾ induced voltage:
¾ Replace N=nxl (l: length of coil):
¾ Note: A x l is just the volume of the coil
¾ So:
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example
¾ A solenoid with 1000 windings is 10 cm long and has an
area of 1cm2. What is its inductance?
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R
L
An RL circuit
I
V
A solenoid and a resistor are placed in series. At t=0 the switch is closed.
One can now set up Kirchhoff’s 2nd law for this system:
If you solve this for I, you will get:
The energy stored in the inductor :E=½LI2
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RL Circuit II
R
L
energy
is released
I
V
energy
is stored
¾ When the switch is closed the current only rises slowly because the
inductance tries to oppose the flow.
¾ Finally, it reaches its maximum value (I=V/R)
¾ When the switch is opened, the current only slowly drops, because the
inductance opposes the reduction
¾
is the time constant (s)
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question
R
L
I
V
¾ What is the voltage over an inductor in an RL circuit long
after the switched has been closed?
¾ a) 0 b) V/R c) L/R d) infinity
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example
R
L
I
V
¾
¾
¾
¾
Given R=10 Ohm and L=2x10-2 H and V=20 V.
a) what is the time constant?
b) what is the maximum current through the system
c) how long does it take to get to 75% of that
current if the switch is closed at t=0
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lon-capa
¾ you should now do questions 8 and 9 of lon-capa set 6.
¾ For question 9, note that the voltage over the inductor is
constant and the situation thus a little different from the
situation of the previous page. You have done this before
for a capacitor as well…
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