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1 2 A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: 1.Current clockwise;

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Presentation on theme: "1 2 A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: 1.Current clockwise;"— Presentation transcript:

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3 2 A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: 1.Current clockwise; force up 2.Current counterclockwise; force up 3.Current clockwise; force down 4.Current counterclockwise; force down example

4 3 The I dl x B force on the coil is a force which is trying to keep the flux through the coil from increasing by slowing it down (Lenz’s Law again). Answer: 3. Current is clockwise; force is down The clockwise current creates a self-field downward, trying to offset the increase of magnetic flux through the coil as it moves upward into stronger fields (Lenz’s Law).

5 4 A rectangular wire loop is pulled thru a uniform B field penetrating its top half, as shown. The induced current and the force and torque on the loop are: v B out 1.Current CW, Force Left, No Torque 2.Current CW, No Force, Torque Rotates CCW 3.Current CCW, Force Left, No Torque 4.Current CCW, No Force, Torque Rotates CCW 5.No current, force or torque example

6 5 The motion does not change the magnetic flux, so Faraday’s Law says there is no induced EMF, or current, or force, or torque. Of course, if we were pulling at all up or down there would be a force to oppose that motion. Answer: 5. No current, force or torque v B out

7 32.1 Self-Inductance When the switch is closed, the current does not immediately reach its maximum value Faraday’s law can be used to describe the effect 6

8 Self-inductance, cont. (a) A current in the coil produces a magnetic field directed to the left. (b) If the current increases, the coil acts as a source of emf directed as shown by the dashed battery. (c) The induced emf in the coil changes its polarity if the current decreases. 7

9 Self-inductance occurs when the changing flux through a circuit arises from the circuit itself – As the current increases, the magnetic flux through a loop due to this current also increases – The increasing flux induces an emf that opposes the current – As the magnitude of the current increases, the rate of increase lessens and hence the induced emf decreases – This opposing emf results in a gradual increase in the current 8

10 Self Inductance Define: Self Inductance An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor. The magnetic flux of an inductor is proportional to the current. For each coil (turn) of the solenoid: Φ per coil = AB Φ sol = N(AB) = NAB = NA(µ 0 NI/ℓ) = (Au 0 N 2 /ℓ)I sol 9

11 Inductance of a Solenoid The magnetic flux through each turn is Therefore, the inductance is This shows that L depends on the geometry of the object 10

12 Inductance Units 11

13 Self-inductance, cont. The self-induced emf is given by Faraday’s law and must be proportional to the time rate of change of the current – L is a proportionality constant called the inductance of the device – The negative sign indicates that a changing current induces an emf in opposition to that change 12

14 Inductor has a large inductance (L) and consist of closely wrapped coil of many turns Inductance can be interpreted as a measure of opposition to the rate of change in the current – Remember resistance R is a measure of opposition to the current As a circuit is completed, the current begins to increase, but the inductor produces an emf that opposes the increasing current – Therefore, the current doesn’t change from 0 to its maximum instantaneously 13

15 Inductance of a Solenoid l N A 14

16 Potential difference across an inductor For the ideal inductor, R = 0, therefore potential difference across the inductor also equals zero, as long as the current is constant. What happens if we increase the current? An inductor is a conducting coil. An inductor in a circuit resists change in current with an induced potential. An inductor stores energy in a magnetic field. The “strength” of an inductor is determined by is “inductance”, represented by the letter L. 15

17 Potential difference across an inductor Increasing the current increases the flux. An induced magnetic field will oppose the increase by pointing to the right. The induced current is opposite the solenoid current. The induced current carries positive charge to the left and establishes a potential difference across the inductor. Induced current Induced field Potential difference 16

18 Potential difference across an inductor The potential difference across the inductor can be found using Faraday’s Law: Where Φ m = Φ per coil Φ sol = N Φ per coil We defined Φ = LI dΦ sol /dt = L |dI/dt| Induced current Induced field Potential difference 17

19 Potential difference across an inductor If the inductor current is decreased, the induced magnetic field, the induced current and the potential difference all change direction. Note that whether you increase or decrease the current, the inductor always “resists” the change with an induced current. 18

20 The sign of potential difference across an inductor ∆V L = -L dI/dt ∆V L decreases in the direction of current flow if current is increasing. ∆V L increases in the direction of current flow if current is decreasing. ∆V L is measured in the direction of current in the circuit 19

21 Comparison of R and L in a simple circuit  =-IR  =-L(  I/  t) L is a measure of opposition to the rate of change in current R is a measure of opposition to the current 20

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26 Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circle they produce a magnetic field that is: A. into the page B. out of the page C. to the left D. to the right E. zero e example 25

27 : How much current flows through the resistor? How much power is dissipated by the resistor?       B = 0.15 T v = 2 m/s 50 cm 3  example 26

28 Inductor, L L  LL i When switch is first closed,  L opposes emf of cell. 27

29 Inductor, L When switch is opened,  L supports emf of cell. L  i LL 28

30 29 Energy Stored in Magnetic Field Energy density = Magnetic Energy per unit volume (J/m 3 ) Energy density = B 2 / (2  0 ) Example: 1 Tesla field Energy density = (1 T) 2 / [8  ·10 -7 T ·m /A] = 3.98 ·10 5 T·A / m = 3.98 ·10 5 T·A ·m / m 2 = 3.98 ·10 5 N / m 2 = 3.98 ·10 5 J / m 3

31 : A coil has an inductance of 3.00 mH and the current changes from A to 1.5 A in a time of s. Find the magnitude of the average induced emf in the coil during this time. example 30

32 example 31

33 What e.m.f. will be induced in a 10 H inductor in which current changes from 10A to 7A in 9x10 -2 s ? Solution: L= 10H, I1= 10A, I2= 7A, dt= 9x10-2s 19x10 s ε= -L dI/dt, = -L (I 2 -I 1 )/dt = -10 (7-10)/ 9x10 -2 = Volt example 32

34 Example The current in a 10 H inductor is decreasing at a steady rate of 5 A/s. If the current is as shown at some instant in time, what is the magnitude and direction of the induced EMF? Magnitude = (10 H)(5 A/s) = 50 V Current is decreasing Induced emf must be in a direction that OPPOSES this change. So, induced emf must be in same direction as current (a) 50 V (b) 50 V i 33

35 Example : Find the inductance of a uniformly wound solenoid having N turns and length. Assume that is much longer than the radius of the windings and that the core of the solenoid is air. We can assume that the interior magnetic field due to the source current is uniform and given by Equation where n = N/ is the n umber of turns per unit length. The magnetic flux through each turn is : where A is the cross-sectional area of the solenoid. Using this expression and Equation we find that : Because N = n, we can express the result in the form : V = volume of the solenoid = A 34

36 Example : Calculating Inductance and emf (a)Calculate the inductance of an air-core solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its cross-sectional area is 4.00 cm 2. (b)Calculate the self-induced emf in the solenoid if the current through it is decreasing at the rate of 50.0A/s. Solution for (a) Using Equation (32.4), we obtain : Solution for (b) Using Equation (32.1) and given that dI/dt = -50.0A/s, we obtain : 35

37 An inductor is made by tightly winding 0.30 mm diameter wire around a 4.0 mm diameter cylinder. What length cylinder has an inductance of 0.01 mH? example 36

38 A 10.0 A current passes through a 10 mH inductor coil. What potential difference is induced across the coil if the current drops to zero in 5 ms? example 37

39 Energy in Inductors and Magnetic Fields A magnetic field stores considerable energy. Therefore, an inductor, which operates by creating a magnetic field, stores energy. Let’s consider how much energy UL is stored in an inductor L carrying current I: 38

40 A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find (a) its inductance and (b) the rate at which current must change through it to produce an emf of 75mV. example 39

41 example 40

42 example 41

43 Example: Current I increases uniformly from 0 to 1 A. in 0.1 seconds. Find the induced voltage across a 50 mH (milli-Henry) inductance. Negative result means that induced EMF is opposed to both di/dt and i. i + i - ELEL Apply:Substitute: 42

44 a) At equilibrium (infinite time) how much energy is stored in the coil? L = 53 mH example E = 12 V 43

45 example 44

46 What is the magnetic energy stored in a3-mH inductor when the current through it is 4 mA? 24 × joule or 2.4 × joule example 45

47 What happens to the energy stored by an inductor when the current through it is doubled? Its energy is quadrupled (i.e. 4 times of original) example 46

48 The current through a 4-mH inductor is varying as follows: I = 2t What will be the induced emf at t = 1 second? ε = -24 mV example 47

49 Example Calculate the inductance of a solenoid with 100 turns, a length of 5.0 cm, and a cross sectional area of 0.30 cm 2. L =  0 N 2 A l L = (4  X T m/A)(100) 2 (3 X m 2 ) (0.05 m) L = 7.5 X H or 7.5  H 48

50 Example The same solenoid is now filled with an iron cores (  = 4000  0 ). Calculate the inductance L = (4000)(7.5 X H) L = H or 30 mH 49

51 Energy stored in a Magnetic Field U B = ½ L I 2 – U B : energy stored in magnetic field – L: inductance in Henrys – I: current in amperes 50

52 Inductance (review) Increasing current in a coil of wire will generate a counter emf which opposes the current.emf Applying the voltage law allows us to see the effect of this emf on the circuit equation.voltage law The fact that the emf always opposes the change in current is an example of Lenz's law.Lenz's law The relation of this counter emf to the current is the origin of the concept of inductance.inductance The inductance of a coil follows from Faraday's law.inductance of a coilFaraday's law 51

53 Inductance (review) Inductance of a coil: For a fixed area and changing current, Faraday's law becomesFaraday's law Since the magnetic field of a solenoid ismagnetic fieldsolenoid then for a long coil the emf is approximated byemf 52

54 Inductance (review) From the definition of inductanceinductance we obtain 53

55 Energy in a Magnetic Field Battery Power Resistor Power Inductor Power The energy stored in the inductor For energy density, consider a solenoid 54


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