Induced Emf Book pg 428-432
Induced EMF Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus:
History of Induction 1819 Oersted: a current flowing in a conductor produces a magnetic field 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
How to create an induced current A bar magnet and a helical coil of wire, connected to an ammeter relative motion between the coil and the magnet creates current Only a changing magnetic field induces an emf in the coil, which leads to an induced current
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 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
Or change Fleming s LHR to the right hand: Thumb points in direction of movement of conductor 1 st finger points in direction of magnetic field 2 nd finger points in direction of induced current
Strength of induced current Depends on: Speed of movement Strength of magnetic flux density Number of turns on coil Area of coil 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
What is the difference between magnetic field and flux linkage? 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.
Find the direction of induced emf A) B) C) D) Note: If the conductor moves in opposite direction, current is in opposite direction If conductor moves in same direction as magnetic field, no current is induced
Motional emf in a moving conductor In the conductor, each charge Q moves within the rod and experiences a force F = qvb RHR to find direction: Q+ moves up Q moves down L RHR b/c electron flow Constant v to the right, Perpendicular to B 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
If the ends of the rod are connected to a conducting rail, current will flow Conventional current: ccw Electron flow: cw
Charge separation 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 E s move to the lower end, leave a +Q charge at the top Charge builds up until F Q = F B QE = QvB E = vb At that point charge stops to flow and F net = 0
To determine the magnitude of induced emf: At the top of the rod, F Q = EQ (electric force) where E = electric field due to separated charges But E = V, where V = voltage PD b/w ends of the rod s = emf s = L = length of rod E = emf L emf L Q emf L = F Q Q and E = F Q Q = QvB F Q = Q emf L Motional emf when v, B and L are perpendicular to each other Emf = vbl
Emf = vbl From here we can see that emf = 0 when v = 0 A greater speed and a stronger magnetic field leads to greater emf for a given rod of length L
Faraday found that the strength of the induced current does also depend on: Strength of magnetic flux density Number of turns on coil Area of coil If a conducting wire is a tightly wound coil of N turns we get emf = NvBL Other equations: R = V emf becomes for induced emf R = I I P = VI becomes for induced emf P = emf I
Example Suppose a rod is moving at a speed of 5.0ms 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. Find a) emf b) induced current c) electric power delivered to the bulb d) energy used by the bulb in 60s A) emf = vbl = 5 x 0.80 x 1.6 = 6.4V B) I = emf R = 6.4 96 = 0.067A c) P = I x emf = 0.067 x 6.4 = 0.43W d) Emf = Pxt = 0.43 x 6.4 = 26J
Whenever an induced emf causes a current to flow, a second magnetic force is created. Using the RHR F is opposing v, the motion of the rod By itself, F would slow down the rod 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, but in opposite direction (F B = ILBsin θ)