1.
Induced emf• Experiment: – When the switch is closed, the ammeter deflects momentarily – There is an induced emf in the secondary coil (producing an induced current) due to the changing magnetic field through this coil• When there is a steady current in the primary coil, the ammeter reads zero current – Steady magnetic fields cannot produce a current
2.
Magnetic Flux• In general, induced emf is actually due to a change in magnetic flux• Magnetic flux is defined similarly to electric flux: Φ B = B⊥ A = BA cos θ – Units of flux are T⋅m2 (or Wb) – Flux can be positive or negative• The value of the magnetic flux through a loop/surface is proportional to the total number of field lines passing through the loop/surface
3.
Faraday’s Law• Faraday’s Law says that the magnitude of the induced emf in a circuit is equal to the time rate of change of the magnetic flux through the circuit• The average emf ε induced in a circuit containing N tightly wound loops is given by: ∆Φ B – Magnetic flux changes by amount ∆ΦB ε = −N ∆t during the time interval ∆t• The minus sign in the equation concerns the sense of the polarity of the induced emf in the loop (and how it makes current flow in the loop) – The direction follows Lenzs law: • The direction of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop • The induced current tends to maintain the original flux through the loop
4.
Example Problem #20.58The square loop shown is madeof wires with a total seriesresistance of 10.0 Ω. It is placedin a uniform 0.100-T magneticfield directed perpendicular intothe plane of the paper. The loop,which is hinged at each corner,is pulled as shown until theseparation between points A andB is 3.00 m. If this process takes0.100 s, what is the averagecurrent generated in the loop?What is the direction of thecurrent? Solution (details given in class): 0.121 A clockwise
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Interactive Example Problem: Flux and Induced emf in a Loop Animation and solution details given in class.(PHYSLET Physics Exploration 29.3, copyright Pearson Prentice Hall, 2004)
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Example Problem #20.29Find the direction of the current in the resistor R shown in the figure after eachof the following steps (taken in the order given): (a) The switch is closed. (b)The variable resistance in series with the battery is decreased. (c) The circuitcontaining resistor R is moved to the left. (d) The switch is opened. Solution (details given in class): (a) Right to left (b) Right to left (c) Left to right (d) Left to right
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Interactive Example Problem: Lenz’s Law Animation and solution details given in class.(PHYSLET Physics Exploration 29.1, copyright Pearson Prentice Hall, 2004)
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Applications of Faraday’s Law• Ground fault interrupter (GFI)• Electric guitar pickups• Magnetoencephalography (brain), magnetocardiograms (heart and surrounding nerves)
9.
Motional emf• Consider a straight conducting bar moving perpendicularly to the direction l of a uniform external magnetic field – Free charges (e–) experience downward force F – Creates a charge separation in the conductor – Charge separation leads to a downward electric field E – Eventually equilibrium is reached and the (downward) magnetic force qvB on the electrons balances the (upward) electric force qE • qE = qvB E = vB • ∆V = El (since E uniform) ∆V = El = Blv – A potential difference is maintained across the conductor as long as there is motion through the field • If motion is reversed, polarity of potential difference is also reversed
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Motional emf• Now consider the same bar as part of a closed conducting path – Since the area of the current loop changes as the bar moves, an induced emf is created due to a changing magnetic flux through the loop – An induced current is created – The direction of the current is counterclockwise (Lenz’s law) – Assume bar moves a distance ∆x in time ∆t • ∆ΦB = B∆A = Bl∆x • From Faraday’s Law (and noting that N = 1): ∆Φ B ∆x ε = = Bl = Blv ∆t ∆t (motional emf)
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Motional emf• Notes on this experiment: – The moving bar acts like a battery with an emf given by the previous formula – Although the current tends to deplete the accumulated charge at each end of the bar, the magnetic force pumps more charge to maintain a constant potential difference – At right is the equivalent circuit diagram – The work done by the force pulling the rod is the source of the electrical energy – This setup is not practical; a more practical use of motional emf is through a rotating coil of wire (electric generators) – Note that a battery can be used to drive a current, which establishes a magnetic field, which can propel the bar
12.
Inductive Magnetoreception in SharksPhysics Today, March 2008
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Example Problem #20.21A car has a vertical radio antenna 1.20 m long. Thecar travels at 65.0 km/h on a horizontal road where theEarth’s magnetic field is 50.0 µT, directed toward thenorth and downwards at an angle of 65.0° below thehorizontal. (a) Specify the direction the car shouldmove in order to generate the maximum motional emfin the antenna, with the top of the antenna positiverelative to the bottom. (b) Calculate the magnitude ofthis induced emf. Solution (details given in class): (a) Toward the east (b) 4.58 × 10–4 V
15.
Electric Generators• Alternating current (AC) generators f = 60 Hz (U.S.) ε = NBAω sin ωt (ω = 2πf ) N = # turns in loop coil B = magnetic field strength A = Area of the loop coil ω = angular rotation speedεmax = NBAω (plane of loopparallel to magnetic field) (θ = ωt)εmin = 0 (plane of loopperpendicular to magnetic field)• Direct current (DC) generators(note similarity to DCmotor – Chap. 19)
16.
Motors and Back emf• Generators convert mechanical energy to electrical energy• Motors convert electrical energy to mechanical energy generators run in reverse• As the coil in the motor rotates, the changing magnetic flux through it induces an emf – This emf acts to reduce the current in the coil (in agreement with Lenz’s law) – For this reason, the emf is called a back emf – Power requirements for starting a motor or running it under heavy loads larger than running under average loads
17.
Self-Inductance• Consider the effect of current flow in a simple circuit with a battery and a resistor – Faraday’s law prevents the current from reaching its maximum value (ε / R) immediately after switch is closed • Magnetic flux through the loop changes as current increases • Resulting induced emf acts to reduce total emf of circuit – When the rate of current change lessens as current increases, induced emf decreases• Another example: increasing and decreasing current in a solenoid coil
18.
Self-Inductance• These effects are called self-induction• The emf that is produced is called self-induced emf (or back emf) ∆I ε = −L ∆t – L is a constant called the inductance – The minus sign indicates that a changing current induces an emf that is in opposition to that change – Units of inductance is the henry (H) 1 H = 1 V⋅s / A• An expression for inductance can be found by equating Faraday’s law with the above: ∆Φ B ΦB L=N =N ∆I I
19.
RL Circuits• A circuit element having a large inductance is called an inductor ∆I• From ε L = − L ∆t we see that an inductor acts as a current stabilizer – The larger the inductance in a circuit, the larger the opposition to the rate of change of the current – Remember that resistance is a measure of the opposition to current• Consider a series circuit consisting of a battery, resistor, and inductor – Current changes most rapidly when switch is first closed • Back emf largest at this point (Series RL circuit)
20.
RL Circuits• Using calculus, we can derive an expression for the current in the circuit as a function of time ε (rise of I= R (1 − e − Rt / L ) current) – Similar to charge q as a function of time for a charging capacitor – Like RC circuits, there is a time constant for the circuit: L τ= R• When the switch is opened, the current in the circuit decreases exponentially with time ε − Rt / L I= e (decay of current R )
21.
Example Problem #20.53An 820-turn wire coil of resistance 24.0 Ω isplaced on top of a 12500-turn, 7.00-cm-longsolenoid as shown. Both coil and solenoid havecross-sectional areas of 1.00×10–4 m2. (a) Howlong does it take the solenoid current to reach0.632 times its maximum value? (b) Determinethe average back emf caused by the self-inductance of the solenoid during this interval.(c) The magnetic field produced by the solenoidat the location of the coil is ½ as strong as thefield at the center of the solenoid. Determine theaverage rate of change in magnetic flux througheach turn of the coil during the stated interval.(d) Find the magnitude of the average Solution (details given in class):induced current in the coil. (a) 20.0 ms (b) 37.9 V (c) 1.52 mV (d) 51.8 mA