Ch. 31 - Electromagnetic Induction

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Transcript Ch. 31 - Electromagnetic Induction

Faraday’s Law of Induction
Electromagnetic Induction
A changing magnetic field (intensity, movement)
will induce an electromotive force (emf)
In a closed electric circuit,
a changing magnetic field
will produce an electric current
Faraday’s Law of Induction
The induced emf in a circuit is proportional to the rate of change
of magnetic flux, through any surface bounded by that circuit.
Ɛ = - B / t
Magnetic Flux

A
B
• In the easiest case, with a constant magnetic field B, and a
flat surface of area A, the magnetic flux is
B = B A cos 
• Units : 1 tesla x m2 = 1 weber
Magnetic Flux
What is the flux of the magnetic field
In each of these three cases?
a)
b)
c)
Find the magnetic flux through the loop
Loop radius r = 2.50 cm,
magnetic field B = 0.625 T.
Faraday’s Experiments
N
I
•
•
•
•
•
S
v
Michael Faraday discovered induction in 1831.
Moving the magnet induces a current I.
Reversing the direction reverses the current.
Moving the loop induces a current.
The induced current is set up by an induced EMF.
Faraday’s Experiments
(right)
(left)
I/t
I
EMF
S
• Changing the current in the right-hand coil induces
a current in the left-hand coil.
• The induced current does not depend on the size of
the current in the right-hand coil.
• The induced current depends on I/t.
Faraday’s Law
N
1)
i
S
i/t
2)
EMF
v
S
i
• Moving the magnet changes the flux B (1).
• Changing the current changes the flux B (2).
• Faraday: changing the flux induces an emf.
Ɛ = - B /t
The emf induced
around a loop
Faraday’s law
equals the rate of change
of the flux through that loop
Lenz’s Law
• Faraday’s law gives the magnitude and direction of the induced
emf, and therefore the direction of any induced current.
• Lenz’s law is a simple way to get the directions straight, with less
effort.
• Lenz’s Law:
The induced emf is directed so that any induced current flow
will oppose the change in magnetic flux (which causes the
induced emf).
• This is easier to use than to say ...
Decreasing magnetic flux  emf creates additional magnetic field
Increasing flux  emf creates opposed magnetic field
Lenz’s Law
B
N
I
B
S
v
If we move the magnet towards the loop
the flux of B will increase.
Lenz’s Law  the current induced in the
loop will generate a field B opposed to B.
Example of Faraday’s Law
Consider a coil of radius 5 cm with N = 250 turns.
A magnetic field B, passing through it,
changes at the rate of B/t = 0.6 T/s.
The total resistance of the coil is 8 W.
What is the induced current ?
B
Use Lenz’s law to determine the
direction of the induced current.
Apply Faraday’s law to find the
emf and then the current.
Example of Faraday’s Law
Lenz’s law: B/t > 0
B
B is increasing, then the upward
flux through the coil is increasing.
I
Induced B
So the induced current will have
a magnetic field whose flux
(and therefore field) is down.
Hence the induced current must be
clockwise when looked at from above.
Use Faraday’s law to get the magnitude of the induced emf and current.
B
B = B A cos 
I
Ɛ= - B /  t
Induced B
The induced EMF is: Ɛ = - B /  t
In terms of B: B = N(BA) = NB (pr2)
Therefore Ɛ = - N (pr2) B /  t
Ɛ = - (250) (p 0.0052)(0.6T/s) = -1.18 V (1V=1Tm2 /s)
Current I = Ɛ/ R = (-1.18V) / (8 W) = - 0.147 A
Motional EMF
Up until now we have considered fixed loops.
The flux through them changed because the
magnetic field changed with time.
Now, try moving the loop in a uniform and
constant magnetic field.
This changes the flux, too.
B points
into screen
x
x
x
x
x
x
x B
x
x
x
R D
x
x
x
x
x
v
x
Motional EMF - Use Faraday’s Law
x
x
x
x
x
x
x B
x
x
x
R D
x
x
x
x
x
v
x
The flux is B = B A = BDx
This changes in time:
B /  t = BD( x/ t) = -B D v
Hence, by Faraday’s law, there is an induced emf and current.
What is the direction of the current?
Lenz’s law: there is less inward flux through the loop. Hence
the induced current gives inward flux.
 So the induced current is clockwise.
x
x
x
x
x
x
x B
x
x
x
x
R D
x
x
x
x
x
v
Motional EMF
Faraday’s Law
Faraday’s Law B/ t = - Ɛ
gives the EMF  Ɛ = BDv
In a circuit with a resistor, this gives
Ɛ = BDv = IR  I = BDv/R
Thus, moving a circuit in a magnetic field
produces an emf exactly like a battery.
This is the principle of an electric generator.
A New Source of EMF
• If we have a conducting loop in a magnetic field, we
can create an EMF (like a battery) by changing the
value of B = B A cos 
• This can be done by changing the area, by changing
the magnetic field, or the angle between them.
• We can use this source of EMF in electrical circuits in
the same way we used batteries.
• Remember we have to do work to move the loop or to
change B, to generate the EMF (Nothing is for free!).
Motional EMF
What happens when the conducting
horizontal bar falls?
Motional EMF
What happens when the conducting
horizontal bar falls?
1. As the bar falls, the flux of B changes,
so a current is induced, and the bulb
lights.
Motional EMF
What happens when the conducting
horizontal bar falls?
1. As the bar falls, the flux of B changes,
so a current is induced, and the bulb
lights.
2. The current on the horizontal bar is
in a magnetic field, so it experiences
a force, opposite to gravity.
Motional EMF
What happens when the conducting
horizontal bar falls?
1. As the bar falls, the flux of B changes,
so a current is induced, and the bulb
lights.
2. The current on the horizontal bar is
in a magnetic field, so it experiences
a force, opposite to gravity.
3. The bar slows down, until eventually
it reaches constant speed (FB = mg)
Eddy Currents
A circulating current is induced
in a sheet of metal when it leaves
a region with a magnetic field.
The effect of the magnetic field
on the induced current is such
that it creates a retarding force
on the moving object.
What is the external force
necessary to move the bar
with constant speed?
The induced current is
I=Ɛ/R
I=BvL/R
The magnetic force on
the current is:
F=BIL
F = B (BvL/R) L
F = B2 v L2 /R
Mechanical Power
to
Electrical Power
The mechanical power used
to move the bar is:
PM = F v = B2 v2 L2 / R
The electrical power
dissipated in the circuit is:
PE = I2 R = (B v L / R)2 R
PE = B2 v2 L2 / R
The mechanical power
is transformed into
electrical power
Given:
Bulb: R = 12 W, P = 5 W
Rod: L = 1.25 m, v = 3.1 m/s
Find:
a) Magnetic field strength
b) External force
Electric Generator
Simple Electric Motor