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Steve Boddeker's 123 Physics Pages;
Ch23 Magnetic Flux and Faraday's Law of Induction
Michael Faraday purposely set out to determine if Oersted's findings can be true in the opposite.
Faraday experimentally showed that a changing magnetic field produced an electric field. This electromotive force due to the electric field in turn applied a force on the free charged in a conductor, inducing current.
Induced Electromotive Force
We should have this experiment in our lab, at least a variation.Coils of wires around an iron bar…increase # of coils, what happens, does current change, does voltage change???(answer at end of chapter)
But what will you definitely see, when the switch is opened and closed, an induced current on the secondary circuit.
As we have been discussing since first day of class…let’s say the red bands are permanent magnets. As the magnet approaches the coil of wire, the magnetic field approaches.
The stationary coil experiences an increasing then decreasing magnetic field, which induces an electric field in the stationary frame of the coil. This emf that the electric field causes the charges to experience a force.
This can be said, at the most simplistic terms, inducing the charges to move…thus current, or induced electricity.
Magnetic Flux
Φ = BA cosθ(remember cosine is for dot products)
Magnetic flux is used in the calculation of the induced emf.
Units are Webers
Example
At a certain location, the Earth's magnetic field has a magnitude of 7×10−5 T and points in a direction that is 60° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 50 cm by 80 cm.
Φ = BA cosθ
Φ = 7e-5(.5).8 cos(90-60)
Φ = 2.4e-5 Webers
Faraday’s Law of Induction
· Faraday’s law: An emf is induced when the magnetic flux through a loop changes with time.
emf = -N ΔΦ / Δt
Magnetic field lines “change” when the magnet is moving with respect to the coil
Example
Consider a single-loop coil whose magnetic flux is given
a. Is the magnitude of the induced emf in this coil greater near t=4s or near t=5s
b. At what times in this plot do you expect the induced emf in the coil to have a maximum magnitude?
c. Estimate the induced emf in the coil at times near t=1s and t=2s
(a) At 4 seconds, this is analogous to the vertical component of velocity of a ball at the top of its path
(b) Same as above 1, 3, 5 seconds
(c) emf1sec = -N ΔΦ / Δt
emf = -1 (2 - -2) / (0.65 – 1.35)
emf = -4Wb / -.7sec
emf = 5.7 Volts
emf2sec = changing directions
emf2sec= 0 volts
Lenz’s Law
An induced current always flows in a direction that opposes the change that caused it
(Choose either top or bottom to verify the above sketches while using the right hand rule)
Wire in Magnetic Field
Drop Rod (drilled out holes) on rigid vertical wires, in a large magnetic field.
We have an existing magnetic field point out of the page (toward you). A conducting rod connects the right and left side wires. As it falls the magnetic flux decreases, inducing a current.
Eddy Currents
Induced currents, called eddy currents, oppose direction of motion.
Example
Mechanical Work and Electrical Energy
ΔΦ = B ΔA
ΔΦ = B v lΔt
v = Δx/Δt
v Δt = Δx
v Δt l = Δx l
v Δt l = ΔA
emf = -N ΔΦ / Δt
emf = -N B v l Δt / Δt
emf = -N B v l(assume 1 coil)
emf = B v l
emf = B v l
emf/Δx = B v l /Δx
Electric field = B v
E = B v
P = F v
P = B2 v2L2 / R
F = I L B
F = ( V / R) L B
F = (emf / R) L B
F = (B v l / R) L B
F =B2 v L2 / R
Example
Generators and Motors
emf = N B v l
emf = N B vmax l
emf = N B ω r sin(ωt) l
emf = N B ω l r sin(ωt)
emf = N B ω A sin(ωt)
emf = N BA ω sin(ωt)
From 121
v = ω r
vmax = ω r sin (ωt) = ω r
An electric motor is exactly the opposite of a generator—it uses the torque on a current loop to create mechanical energy.
Inductance
Inductance is the proportionality constant that tells us how much emf will be induced for a given rate of change in current:
Defined as
emf = L ΔI / Δt
Then solve for inductance
L = emf (Δt) / ΔI Given emf = N ΔΦ / Δt
L = N ΔΦ / ΔI
Units henry 1 V sec/A = 1 H
For Solenoids
L = µo n2 A l
Example
RL Circuits
· When the switch is closed, the current immediately starts to increase. The back emf in the inductor is large, as the current is changing rapidly. As time goes on, the current increases more slowly, and the potential difference across the inductor decreases.
τ = L / R
Example
Energy Stored in a Magnetic Field
L = emf (Δt) / ΔI (from above)
emf = N ΔΦ / Δt
emf = L ΔI / Δt
Pave = Iave V
Pave = ½ I LΔI/Δt
Pave = ½ I2 L / T
P = Work / time
Pave = U / T
½I2L/T = U / T
U = ½ L I2; remember from capacitors (U = ½CV2)
This circuit has a battery and an inductor, the current in the inductor is increasing with respect to time, where τ = L / R. So what is the average current while charging? Iave = ½ I
, for solenoids L = µo n2 A l
U = ½ L I2
U = ½ µo n2 A l I2
And we know from Ampere’s law B = µo n I
Thus U = ½ B2 A l /µo
We know the volume inside the solenoid is Vol = A length
So Energy/volume µB = B2 / 2µo
Example
Transformers
I2
=
V1
=
N1
I1
V2
N2
P = I V or V = P / I
I2
=
P1 / I1
=
N1
I1
P2 / I2
N2
But energy is conserved, P1 = P2
1 / I1
=
N1
1 / I2
N2
Example
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