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Faraday’s Law
r
The loop of wire shown below has a radius of 0.2 m, and is in a magnetic field that is increasing at a rate of 0.5 T/s.
Since the area of the loop is constant, Δ(BA) = AΔB
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I Have Proof!
If area doesn’t change, but B does…
Therefore, when A is unchanged,
If area changed, but field remained constant, then you would end up with Δ(BA) = BΔA
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Whiteboard Warmup!
r
The loop of wire shown below has a radius of 0.2 m, and is in a 0.8 T magnetic field. The loop is rotated by 90° in 2 seconds along the dotted line shown. The loop contains a 3 Ω resistor.
a) What is the average induced emf?b) What is the induced current, and which way does it flow?
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a) ε = 0.05 Voltsb) I = ε/R = 0.016 A counterclockwise
Since B and A are constant
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Reverse Rail Gun!A metal rod is pulled to the right at constant velocity, along
two conducting rails shown below. What is the direction of the induced current through the circuit as a result?
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Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
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Whiteboard: What will be εinduced?
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!
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Motional EMF
When the amount of a loop’s area that in a magnetic field is changing at a constant rate, Faraday’s Law gives the result
• L is the side that is not changing.• v is the rate of change of the side
that is entering or leaving the field.
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Using Motional EMF
In terms of the quantities shown above, write an expression for
a)The current through the resistor, and direction of current.b)The power output of the resistor in the circuit.c)The force required to pull the metal bar at constant velocity.
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ε = BLv
I = ε/R
a) What is the current through the resistor?
I = BLv/R
Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
II
II
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P = I2R
b) What is the power output of the resistor?
P = (BLv)2/R
P = B2L2v2/R
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c) What is the force required to pull the bar at constant speed?
I
Using RHR #2, you can determine that when there is a current flowing through the circuit, the moving
metal bar will feel a magnetic force to the left.
FB
Therefore, to pull the bar at constant velocity, you must exactly balance out the magnetic force BIL.
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c) What is the force required to pull the bar at constant speed?
IFB
FB = BIL
FB = B(BLv/R)L
FB = B2L2v/R
In order to pull the bar at constant velocity, you must exactly match this force by pulling to the right
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Let’s analyze the results for a second…
Presistor = B2L2v2/R F = B2L2v/R
P = Force x velocity!The power output of the resistor will be exactly equal to the
power delivered to the system by pulling the rod.
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Coils with Multiple Loops!
Each coil acts as its own loop.
If there are N coils,
Just multiply by N!
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Solenoid
Φ = N*BA
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Solenoids are useful!
They multiply the magnetic flux, and therefore the induced emf, by the number of turns that the wire has
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Ring Launcher!
I
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Ring Launcher!
I
Bcoil
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This induces a current in the ring that opposes the field of the coil
I
Bcoil
XS
XS
XS
XS
Bring
Iinduced
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I
The current-carrying coil of wire acts like a magnet, with the field lines coming out of North and into South.
N
S
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XS
XS
XS
XSBring
Iinduced
XS
XS
XS
XS
NS
The current-carrying ring also acts like a magnet, with the field lines coming out of North and into South.
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XS
XS
The net result looks like this!
I
BcoilXS
XS
XS
XS
Bring
Iinduced NS
XS
XS
NS
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Strong repulsion!!!
I
Iinduced NS
NS
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Whiteboard: Copper Tube Drop!
N
S
v
a) What will be the direction of the induced current in each of these sections of copper tube?
b) Draw the “magnet” that each of these sections acts like.
c) What will be the result when the magnet is dropped down the tube?
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v
Φ: DownwardΔΦ: UpwardBinduced: Downward
Φ: DownwardΔΦ: UpwardBinduced: Downward
Iind
Φ: DownwardΔΦ: DownwardBinduced: Upward
Φ: DownwardΔΦ: DownwardBinduced: UpwardIind
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v v
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v
Attracted by the induced magnet above
Repelled from the induced magnet below
The magnet will fall slowly!!!