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Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
Ch30 induction r
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Ch30 induction r

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  • 1. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Ch 30: Induction and Inductance Sections 1-11 3.5 lectures We started magnetism by looking at force on a charge or a wire caused by a magnetic field - defined a magnetic field by: Then we looked at creation of magnetic fields by currents and Biot-Savart Law and Ampere’s Law and found force between two wires or the magnetic field in a solenoid. This chapter studies current or emf induced in a solenoid or coil by a changing magnetic field, including changing magnetic field generated by current in the first place (i.e. inductance). Ch30-1/66 An experiment Move a bar magnet thru a coil and measure the current. -a current appears when magnet is in motion relative to loop -faster motion produces greater current -if moving N pole into loop causes current in one direction, then when moving out, current is reversed - moving S pole into the loop causes the reverse current to flow compared to N pole into loop. Current is induced current. Work done per unit charge to produce current is induced emf, -process of producing current is induction. Ch30-2/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 1
  • 2. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Another experiment Two conducting loops. Close but not touching. Close S to cause current flow in right loop. => briefly see an induced current on meter. Briefly see an induced current again when we open S, but in opposite direction. There is no current in left-hand loop when there is a constant current in right. Ch30-3/66 Faraday’s Law of induction Michael Faraday (as in farad) generalized these experimental results by realizing that the current was caused in a loop when the number of magnetic field lines passing thru loop is changed. Critical issue is rate of change of number of lines, not number of lines (however, the stronger the field, the more likely that rate of change is large). This observation does not explain why it happens, just that it does (just as Coulomb’s Law doesn’t explain why). Ch30-4/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 2
  • 3. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Magnetic flux electric flux Gauss’ Law Here, in analogy we define magnetic flux where, as before, dA is a vector of magnitude dA, normal to a differential area dA. Unit is tesla-sq meter, or weber We talk of flux thru a loop by doing integral over area in a loop. For a flat loop and uniform B | loop, we have Ch30-5/66 Faraday’s Law of induction (cont) Magnitude of emf induced in a conducting loop is equal to rate of change of magnetic flux with time where - sign tells us emf opposes change in ΦB. For a coil with N turns: Check units Ch30-6/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 3
  • 4. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Faraday’s Law of induction (cont) How to change magnetic flux in a loop: -change magnitude of B -change area of coil in magnetic field -change angle of coil wrt to B (B.dA term) Ch30-7/66 Question Given B(t) of a magnetic field which is otherwise uniform throughout a conducting loop and | loop. Rank 5 regions, largest to smallest in terms of induced emf in loop Largest: Smallest: Ch30-8/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 4
  • 5. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Given B(t) of a magnetic field which is otherwise uniform throughout a conducting loop and | loop. Rank 5 regions, largest to smallest in terms of induced emf in loop Largest: Ans: (b) since rate of change highest Smallest: Ans: (a) and (c) since rate of change = 0 and (d) & (e) are same, value in middle. Ch30-9/66 Sample problem A long solenoid S, 220 turns/cm, i = 1.5 A, diameter D is 3.2 cm. At centre a 130-turn coil, C, diameter d = 2.1 cm. Current in S reduced to 0 at constant rate in 25 ms. What is magnitude of induced emf in C while current is changing? Key ideas: There is a magnetic flux in C. It changes and so Faraday’s Law applies where N is 130. Since current reduces constantly Ch30-10/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 5
  • 6. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) Key ideas: Calculate by evaluating the initial ΦB. B is uniform along the axis and thus | area where we used eqn for B in a solenoid. Final flux is 0. One ignores the - sign since only asked for magnitude. Ch30-11/66 Lenz’s Law An induced current has a direction such that magnetic field due to current opposes change in magnetic flux that induces current. Also, direction of an induced emf is that of induced current. Opposition to pole movement: Induced current creates a magnetic dipole of loop (directed from south to north). To oppose incoming N pole => induced N-pole is to right. By RHR => current is as shown. When we pull magnet out, it is in reverse direction so S-pole is trying to pull bar magnets N pole as it leaves. Ch30-12/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 6
  • 7. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Lenz’s Law (cont) An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current. Also, the direction of an induced emf is that of the induced current. Opposition to flux change: With bar magnet away, initially there is no magnetic flux thru loop. As N-pole approaches, B is directed to left (leaves the N-pole). So flux changing and a current is induced to create a magnetic field in opposition, i.e. to right. Hence current induced as shown in (a), again following RHR. Note opposing change in B does not mean in induce Bi is always in opposite direction, eg (b) as we pull magnet out. Ch30-13/66 Electric guitars The permanent magnet (not an electro-magnet from coil) produces a north-south pole in section of wire above it. When this little magnet vibrates, there is a current induced in coil which alternates at same frequency as string and this is amplified. Typically there are more than 1 set of pickups which emphasize various frequencies. (really more complex: bar magnet is inducing currents in string which create a magnetic field which then induces current in coil). Ch30-14/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 7
  • 8. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Three identical loops in uniform magnetic fields which are either increasing or decreasing at identical rates in each half. Which has largest induced current? Smallest? Ch30-15/66 Question Three identical loops in uniform magnetic fields which are either increasing or decreasing at identical rates in each half. Which has largest induced current? Ans: (a) and (b) are same and largest (note different directions) Smallest? Ans: (c) has no net change Ch30-16/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 8
  • 9. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem Uniform magnetic field out of the screen. r= 20cm. Resistance = 2.0 Ω Ideal battery, emf = 2.0 V B = 4.0 t2 + 2.0 t + 3.0 What is magnitude and direction of emf induced by B at t=10 s? Key ideas: Faraday’s law of induction applies and because field | to loop, ΦB = B A and A unchanging: Direction: Flux is out of screen and increasing, so induced current must create a magnetic field opposing this (Lenz’s Law) and so is directed into the screen. So RHR tells us the current is clockwise, as is the induced emf (which opposes battery). Ch30-17/66 Sample problem(cont) Uniform magnetic field out of the screen. r= 20 cm. Resistance = 2.0 Ω Ideal battery, emf = 2.0 V B = 4.0 t2 + 2.0 t + 3.0 What is the current at t = 10 s? Key ideas: The induced emf 5.2 V is greater than the battery emf hence and direction is clockwise as in previous part. Ch30-18/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 9
  • 10. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Another sample problem Rectangular loop. B into screen and varies with t and x. What is magnitude and direction of induced emf around the loop at 0.10s? Key ideas: Since B is | loop, but varies with x, we write (with dA=Hdx): and then The trick is write dA as shown so that integral is simple since B is only varying with x. If it varied with y too, we would need a double integral (dx dy). Direction: B is increasing into screen => current is counterclockwise by RHR so induced B is outwards. Ch30-19/66 Induction and energy transfers As we move the magnet in or out of the loop, the induced magnetic force resists it => hand does work. The induced current heats the wire. So the applied work ends up as thermal energy (ignoring, for now, the radiated energy of em waves). If we move more quickly, the induced current and hence field is greater => we do more work. The current always transfers thermal energy because of the electrical resistance (except in superconductors). Ch30-20/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 10
  • 11. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Induction and energy transfers (cont) One end of a rectangular loop moving relative to a uniform magnetic field into screen with constant v to the right. How much work must the hand do? As the loop moves, the magnetic flux in the loop (B.dA) changes since dA is changing and this induces a current in the loop (Faraday’s law). The magnetic field then exerts a force on the current in the wire and the power required by the hand is P = Fv. Step 1: calculate the current induced. Faraday’s Law gives: accounting for resistance in the loop => Ch30-21/66 Induction and energy transfers (cont) Direction of current must be clockwise to create a current into the screen to oppose the decrease in the magnetic flux. Step 2: Find forces using the general relationship for force acting on wires in magnetic fields. RHR => directions as shown and F2, F3 are equal and opposite => net 0, leaving F1 which is directed opposite to force that hand applies. From above eqn: The net force is 0 => no acceleration => v is constant and rate of work by the hand is: Ch30-22/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 11
  • 12. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Induction and energy transfers (cont) Step 3: Find rate of thermal energy in the loop i.e. the same as the work done by the hand. The work done moving the loop appears as thermal energy in the loop. Ch30-23/66 Eddy currents If we have a conductor in a changing magnetic flux, currents will be induced which create a magnetic field to oppose the change in magnetic field. The force on this current will be to oppose the change in magnetic flux => it will tend to stop the motion (as in previous example) Causes heating in the conductor. Pivot case: As enters field, current counterclockwise to create opposing magnetic field out of screen => force to left => slows it. Once all in magnetic field, no net force and reverses as leaves. Ch30-24/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 12
  • 13. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Eddy currents (cont) In a solid conductor (top, right) if the magnetic field changes, there is a large loop/area defined => large eddy current possible. Laminate the conductor (insulate between layers) => much smaller areas => smaller changes in magnetic flux => smaller currents and power losses. Magnetic Brakes Rotating conducting plate. Turn on magnetic field induces eddy currents which create a force opposing the motion. Steady magnetic field is constantly changing location=> brake stays on (until stopped!): Advantage is lack of wear. Ch30-25/66 Question 4 wire loops with edge lengths of L or 2 L, moving with constant velocity thru a uniform magnetic field B out of the screen. Rank them in terms of maximum induced emf? Largest? Smallest? Ch30-26/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 13
  • 14. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question 4 wire loops with edge lengths of L or 2 L, moving with constant velocity thru a uniform magnetic field B out of the screen. Rank them in terms of maximum induced emf? Ans: The change in ΦB is proportional to Largest? dA/dt which is proportional to the length of the loop up/down the screen. Thus c & d are the same and twice as large Smallest? as a & b. There will be a difference in how long the emf is present, but not the magnitude in the two cases Ch30-27/66 Electric generators (not in text) Consider a wire loop with N turns rotating in a constant magnetic field (as in motor). θ is angle between normal to loop and magnetic field If loop to move with constant angular velocity, so θ = ω t Faraday’s Law gives: Current thru resistance R: Ch30-28/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 14
  • 15. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Electric generators (cont) http://www.walter-fendt.de/ph14e/generator_e.htm which is available off my course web page www.physics.carleton.ca/~drogers/phy1004 “I cannot understand why there is an energy crisis at all. Elementary physics shows that if we move a conductor around a magnet we can produce as much electricity as we need. It is ridiculous that this infinite source of energy has never been used” Paraphrase of a letter to the Ottawa Citizen Why should the author feel like a fool? Ch30-29/66 Generators (cont) Consider power delivered to a resistance R by current generated. Now consider power delivered to generator by torque required to turn loop. i.e. there is no free lunch - or free electrical power Ch30-30/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 15
  • 16. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Induced electric fields Consider a copper ring, radius r, in a uniform magnetic field into screen of radius R. Assume B changing at a uniform rate, hence ΦB thru loop is changing. Faraday’s law tells us there is an induced current in ring and Lenz’s Law and RHR tell us it must be counterclockwise to generate an opposing field out of screen. There must be an electric field which is moving electrons. A changing magnetic field produces an electric field which is called the induced electric field. This field exists even if there is no copper wire and no current! Ch30-31/66 Induced electric fields Consider a circular path, radius r. The electric field must be tangent to the circle by symmetry => circle is an electric field line. There must be concentric circular field lines at any radius (as long as magnetic field is changing). They would change direction if rate of change of magnetic field changed direction. Ch30-32/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 16
  • 17. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Reformulation of Faraday’s Law Consider qo moving around the circular path at r. Work done in one revolution is where is the induced emf, i.e. the work done per unit charge to move a test charge around the loop. Also we have Setting the expressions for W equal each other More generally and hence Ch30-33/66 Reformulation of Faraday’s Law (cont) This eqn allows a new meaning of Previously, induced emf meant work per unit charge done to maintain a current induced by a changing magnetic field or induced emf meant the work done per unit charge on a test charge that moves around a closed path in a changing magnetic field. The above equation tells us that the induced emf is the integral of the above dot product around any closed loop where E is the electric field induced by the changing magnetic field Ch30-34/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 17
  • 18. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Reformulation of Faraday’s Law (cont) Combine the above two equations to give: This is a more general formulation of Faraday’s Law. It tells us that a changing magnetic field induces an electric field. Applies to any closed path in a changing magnetic field. Ch30-35/66 A new look at electric potential We have discussed electric fields produced 2 ways. By static charges and by changing magnetic fields. Both electric field exert forces on charges. Field lines from induced electric fields form closed loops whereas field lines from static charges start on +ve charges and stop on -ve charges. Electric potential has meaning only for electric fields produced by static charges; it has no meaning for electric fields produced by induction Ch30-36/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 18
  • 19. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem Uniform magnetic field into screen. R = 8.5 cm dB/dt = 0.13 T/s. Find E(r) for r<=R Apply Faraday’s Law: Use a circular path of integration & fact that E constant by symmetry. Look at left hand side: RHS: first magnetic flux: Applying Faraday’s law (less the minus sign) Plugging in values at r=5.2 cm Ch30-37/66 Sample problem (cont) Find E(r) for r>=R Apply Faraday’s Law again. RHS: first magnetic flux since B=0 for r>R: LHS: same as before Applying Faraday’s law (less the minus sign) and solving for E Note they match at r=R The changing magnetic field induces an electric field even outside changing magnetic field. Ch30-38/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 19
  • 20. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Inductors and inductance Just as capacitors created electric fields and we took a parallel plate capacitor which created a uniform electric field as a basic type, a device called an inductor (symbol ) can be used to create a desired magnetic field. A long solenoid, or a short section near the middle of a solenoid, is taken as the basic type of inductor. We define the inductance of inductor as where ΦB is the magnetic flux created in the inductor by the current i and N is the number of turns. The windings are linked by shared flux. Refer to NΦB as magnetic flux linkage. Ch30-39/66 Inductors and inductance (cont) SI unit of inductance is the henry (H) There is an assumption in chapter 30 that there is no iron in/near the inductors. Ch30-40/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 20
  • 21. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Inductance of a solenoid Consider a long solenoid, cross section A, n turns per unit length. What is the inductance per unit length near the centre? Start by calculating the magnetic flux linkage NΦB . For a length l near the centre: where B is magnitude of the magnetic field in the solenoid and from before we have: Hence, from definition of L we get: and hence Ch30-41/66 Inductance of a solenoid (cont) The inductance depends only on the geometry of the device, not the current or the magnetic field. (like a capacitor) Note the n2 dependence: if you triple the number of turns per unit length, this increases the value of N by 3 and also increases the B by a factor 3 and hence the flux linkage NΦB and the inductance increase by a factor of 9. For a solenoid with L>>r, the eqn for the overall L is accurate, but it ignores end effects. We can rewrite the units for based on these equations. Ch30-42/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 21
  • 22. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Self-induction We have seen that if two coils (inductors) are near each other, a current in the one coil produces a magnetic flux through the second coil. If the current changes, then the magnetic flux changes and hence an emf is induced in the second coil according to Faraday’s law. There is an induced emf in the first coil as well. An induced emf appears in any coil in which the current is changing From the definition of inductance: and Faraday’s Law tells us: Hence: i.e. self-induced emf appears whenever current is changing Ch30-43/66 Self-induction(cont) Note that the magnitude of the current does not affect the induced emf directly, only the rate of change does. Lenz’s law tells us that the direction of the induced emf opposes the change in the current. We can drop the sign if we only want the magnitude of the induced emf. Ch30-44/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 22
  • 23. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Self-induction(cont) Assume di/dt is +ve. Lenz’s law says the the induced emf will oppose the increasing current. Direction will be as shown - or conversely if current decreasing. We cannot define an electric potential within the inductor (as discussed previously) BUT we can define potentials at points in the circuit which are not in the inductor. A self-induced potential difference, VL, exists across an inductor (say across terminals outside the changing magnetic flux). For ideal inductor VL = induced emf, otherwise it is like the potential across a real battery (emf - iR). Ch30-45/66 Question Figure shows an induced emf induced by the current thru the coil. Which statement describes the current in the coil? (a) constant and rightward (b) constant and leftward (c) increasing and rightward (d) decreasing and rightward (e) increasing and leftward (f) decreasing and leftward Ch30-46/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 23
  • 24. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Figure shows an induced emf induced by the current thru the coil. Which statement describes the current in the coil? (a) constant and rightward (b) constant and leftward (c) increasing and rightward (d) decreasing and rightward (e) increasing and leftward (f) decreasing and leftward Ans: induced emf will cause current to the right so (e) since induced current opposes it, or (d) for same reason Ch30-47/66 RL circuits Previously, we saw that if an emf was switched into an RC circuit at t=0, then: where and similarly, when switched off: i.e. it takes time for charge to build or decay. In same way, current rises or falls in a circuit with a resistance R and an inductor L. Without L, current would become emf/R, but with inductor, self-induced emf opposes current at first (Lenz’s Law). Current rises in time to emf/R. When the emf is first switched on self-induced emf is enough to initially cause current to be zero. So initially inductor opposes change in current through it, but in time it acts like an ordinary wire. Ch30-48/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 24
  • 25. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RL circuits (cont) With the switch at (a), the circuit is as at right. Apply the loop rule: Resistance: -iR potential drop across resistor Inductor: induced emf magnitude is L di/dt upward to oppose current flowing as shown =>from y to z the potential drop is -L di/dt. Battery: +emf Hence: or This has same form as RC circuit Ch30-49/66 RL circuits (cont) So in analogy to the RC circuit, with i replacing q etc, the solution is: where the inductive time constant is: For t = 0, initial current is: ? For current is:? VR = iR VL = L di/dt Ch30-50/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 25
  • 26. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RL circuits (cont) So in analogy to the RC circuit, with i replacing q etc, the solution is: where the inductive time constant is: For t = 0, initial current is: ? Ans: i(0) = 0 since e-0 = 1 For current is:? Ans: i( ) = VR = iR VL = L di/dt Ch30-51/66 Inductive time constant What are the units of the inductive time constant? i.e. after 1 time constant current reaches 63% of its equilibrium value. Ch30-52/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 26
  • 27. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RL circuits (cont) Take the battery out of the circuit: (throw switch to b and then break circuit at a using a make before break switch.) compare to the analogous RC circuit So here soln is where we have assume the switch was at a long enough to reach the equilibrium current. Ch30-53/66 Sample problem Circuit with 3 identical resistors of 9 ohms and 2 identical inductors with inductance L = 2.0 mH and an ideal battery with emf of 18 V. (a) What is the current thru the battery just after the switch is closed? Key idea: Just after a switch is closed, the inductors oppose the flow of current, and in fact look like an open circuit for a moment after the switch is closed (as in (b)). So we have a simple single loop circuit Ch30-54/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 27
  • 28. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) (b) What is the current thru the battery a long time after the switch is closed Key idea: A long time after, the inductances offer no resistance to the current flow and hence the circuit is as in (c), 3 parallel resistors. Ch30-55/66 Question Three circuits with identical batteries, inductors and resistors. Rank them according to the current through the battery (a) just after the switch is closed (b) a long time later Ch30-56/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 28
  • 29. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Three circuits with identical batteries, inductors and resistors. Rank them according to the current through the battery (a) just after the switch is closed (b) a long time later (a) Ans: (1) is 0, (2) is emf/R, (3) is emf/2R => 2 is greatest (b) Ans: (1) is emf/2R, (2) is 2emf/R, (3) is emf/R => 2 greatest Ch30-57/66 Energy stored in a magnetic field For an LR circuit the loop rule gives: and multiplying both sides by i gives: Term 1: if dq passes thru battery with emf in time dt, work done is and so rate of work by battery is or : i.e. term 1 is rate at which the emf delivers energy to the rest of circuit Term 3: rate at which thermal energy appears in resistor Term 2: by conservation of energy, this must be the rate at which energy is stored in the magnetic field, i.e. Ch30-58/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 29
  • 30. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Energy stored in a magnetic field(cont) Rewrite as: And integrating to get the total energy stored gives: Hence: This is the total energy stored by an inductor with inductance L carrying a current i. cf the electric field case for the energy stored by a capacitor with capacitance C: Ch30-59/66 Sample problem A coil has inductance, L = 53 mH and resistance 0.35 Ω. If a 12 V emf is applied across the coil, how much energy is stored in the magnetic field when equilibrium is reached? Key idea: and hence we need i at equilibrium. We know so Hence: Part (b) How many time constants are needed to store 1/2 of this energy in the magnetic field? Ch30-60/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 30
  • 31. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) Part (b) How many time constants are needed to store 1/2 of this energy in the magnetic field? Key idea: We want which means: hence: => rearranging gives: i.e. it has stored 1/2 of the equilibrium stored energy after 1.23 time constants (when it has reached 0.71 of the current) Ch30-61/66 Energy density of a magnetic field Consider a length l near the middle of a long solenoid of cross-sectional area A carrying current i. Vol = Al. Since the field outside the solenoid is zero and it is uniform within the solenoid, then the energy stored per unit volume of the field is: where we use: Substituting: Although derived for a special case, this is generally true and corresponds to the similar equation for E: Ch30-62/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 31
  • 32. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Given the following about 3 solenoids, rank them according to the energy density within them, greatest first. Solenoid turns/length current Area a 2n i 2A b n 2i A c n i 6A Ch30-63/66 Question Given the following about 3 solenoids, rank them according to the energy density within them, greatest first. Solenoid turns/length current Area a 2n i 2A b n 2i A c n i 6A Ans: a: 4 n2 i2 b: 4 n2 i2 c: n2 i2 so a = b > c Ch30-64/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 32
  • 33. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem Long coaxial cable: two thin walled concentric conducting cylinders, radii a, b. Inner carries current i, outer carries return path for current. Current sets up a magnetic field. Calculate energy stored in magnetic field for length l of cable Key ideas: Calculate UB from uB using -uB depends on B which we determine from Ampere’s Law given current i. Step 1- Find B: Ampere’s Law for Amperian loop between cylinders: Note, outside second cylinder, total enclosed current is net 0 because in opposite directions => B= 0 outside. Ch30-65/66 Sample problem(cont) Step 2 -find uB using B we just found Step 3 -find UB by integrating over volume Ch30-66/66Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 33

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