Book pg 428 - 432
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Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus
:

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1819 Oersted: a current flowing in a conductor produces a magnetic field
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11 years later Faraday and Henry showed:
A moving magnetic field produces a current in a conductor
Induced current is produced provided there is a change in magnetic flux
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A bar magnet and a helical coil of wire, connected to an ammeter
 relative motion between the coil and the magnet creates current
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Only a changing magnetic field induces an emf in the coil, which leads to an induced current
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If the conductor moves across a magnetic field, a current is also induced movement
This should be a left hand,
“For visualization only”
Movement of conductor
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To find the direction of induced current, change RHR to left hand:
Palm of hand points in direction of motion of conductor
Fingers point in direction of magnetic field
Thumb gives direction of induced current
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Or change Fleming’s LHR to the right hand:
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Thumb points in direction of movement of conductor
1 st finger points in direction of magnetic field
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2 nd finger points in direction of induced current
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Depends on:
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Speed of movement
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Strength of magnetic flux density
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Number of turns on coil
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Area of coil
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The magnitude of induced emf is proportional to the rate of change of magnetic flux Φ with flux linkage
NΦ
Flux linkage NΦ is NOT proportional to the rate of change of magnetic field B
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The magnetic flux density is a measure of the strength of a magnetic field along a particular direction at a particular point.
If you measure the flux density perpendicular to the magnetic field then the flux density is zero in that direction (even though the magnetic field is not).
On the other hand, if the surface is perpendicular to the field then there is magnetic flux passing through the surface
If the surface is parallel to the field then there is no magnetic flux passing through the surface.
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A) B)
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C) D)
Note:
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If the conductor moves in opposite direction, current is in opposite direction
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If conductor moves in same direction as magnetic field, no current is induced
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In the conductor, each charge Q moves within the rod and experiences a force F = qvB
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RHR to find direction:
Q+ moves up
Q – moves down
RHR b/c electron flow
Constant v to the right,
Perpendicular to B
L
Positive and negative charge accumulate until 𝐹
𝑄
𝑜𝑓 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = 𝐹
𝐵
EQLB is reached, no more charge separation
In the rod an emf is induced, which results from the motion of charges through the field
Emf exists as long as the rod is moving

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If the ends of the rod are connected to a conducting rail, current will flow
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Conventional current: ccw
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Electron flow: cw
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The electrons in the conductor experience a force
𝐹
𝐵
= 𝑄𝑣𝐵
From Newton’s 2 nd law: F = ma
Electrons accelerate in response to the force
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E’s move to the lower end, leave a +Q charge at the top
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Charge builds up until 𝐹
𝑄
= 𝐹
𝐵
𝑄𝐸 = 𝑄𝑣𝐵
𝐸 = 𝑣𝐵
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At that point charge stops to flow and 𝐹 𝑛𝑒𝑡
= 0
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At the top of the rod, 𝐹
𝑄
= 𝐸𝑄 (electric force) where E = electric field due to separated charges
But 𝐸 =
𝑉 𝑠
, where V = voltage PD b/w ends of the rod
= emf
s = L = length of rod
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𝐸 = 𝑒𝑚𝑓
𝐿 𝑒𝑚𝑓
𝐿
𝑎𝑛𝑑 𝐸 =
=
𝑄×𝑒𝑚𝑓
𝐿
𝐹
𝑄
𝑄
 𝐹
𝑄
= 𝑄𝑣𝐵
Emf = vBL
=
𝐹
𝑄
𝑄
𝑄×𝑒𝑚𝑓
𝐿
Motional emf when v,
B and L are perpendicular to each other
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From here we can see that emf = 0 when v = 0
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A greater speed and a stronger magnetic field leads to greater emf for a given rod of length L
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Strength of magnetic flux density
Number of turns on coil
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Area of coil
If a conducting wire is a tightly wound coil of
N turns we get
 𝑒𝑚𝑓 = 𝑁𝑣𝐵𝐿
Other equations:
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𝑅 =
𝑉
𝐼
𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑓𝑜𝑟 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑅 = 𝑒𝑚𝑓
𝐼
𝑃 = 𝑉𝐼 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑓𝑜𝑟 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑃 = 𝑒𝑚𝑓 × 𝐼

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Suppose a rod is moving at a speed of 5.0
𝑚𝑠 −1 in the direction perpendicular to B = 0.80T [into page].The rod has a length of 1.6m and R is neglible. A light bulb has a resistance of 96Ω.
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Find a) emf b) induced current c) electric power delivered to the bulb d) energy used by the bulb in 60s
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A) emf = vBL = 5 x 0.80 x 1.6 = 6.4V 𝑒𝑚𝑓 6.4
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B) 𝐼 = = = 0.067𝐴 c) P = I x emf = 0.067 x 6.4 = 0.43W
𝑅 96 d) Emf = Pxt = 0.43 x 6.4
= 26J
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Whenever an induced emf causes a current to flow, a second magnetic force is created.
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Using the RHR
F is opposing v, the motion of the rod
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By itself, F would slow down the rod
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To keep the rod moving to the right with constant speed v, a counterbalancing force must be applied by an external source such as a hand
This force must be equal to 𝐹
𝐵 direction ( 𝐹
𝐵
= 𝐼𝐿𝐵𝑠𝑖𝑛 𝜃)
, but in opposite
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