PHY 102 Lecture 7 7 1 Induced EMF

PHY 102: Lecture 7 • • • 7. 1 Induced EMF, Induced Current 7. 2 Motional EMF 7. 3 Magnetic Flux 7. 4 Lenz’s Law 7. 5 Mutual and Self Inductance

PHY 102: Lecture 7 Electromagnetic Induction 7. 1 Induced EMF, Induced Current

Basics of This Chapter • Changing magnetic field and/or area produces induced emf • Conductor moving through uniform field produces induced emf • Induced current produces secondary magnetic field

Induced Current • When there is no relative motion between the coil of wire and the bar magnet, no current in coil • A current is created in the coil when the magnet moves toward the coil • A current also exists when the magnet moves away from the coil, but the direction is reversed

Induced EMF • Coil behaves as if it were a source of emf • This is known as the induced emf • This emf causes the induced current

Another Way of Inducing EMF/Current • Another way of inducing an emf is to keep the magnetic field constant and to change the area of the coil • The phenomenon of producing an induced emf with the aid of a magnetic field is called electromagnetic induction

PHY 102: Lecture 7 Electromagnetic Induction 7. 2 Motional EMF

EMF Induced in Moving Conductor-1 • When a conducting rod moves through a constant magnetic field, EMF is induced in rod • This is result of magnetic force that acts on a moving charge • A metal rod of length L moving to the right • The velocity v of the rod is constant and is perpendicular to a uniform magnetic field B • Each charge |q| within the rod also moves with a velocity v and experiences a magnetic force of magnitude F = |q|v. B

EMF Induced in Moving Conductor-2 • Free electrons are driven to the bottom of the rod • This leaves behind an equal amount of positive charge at the top (RHR-1) • “+“ and “-” charges accumulate until attractive electric force that they exert on each other becomes equal in magnitude to the magnetic force • When the two forces balance, no further charge separation occurs

EMF Induced in Moving Conductor-3 • The separated charges on the ends of the moving conductor give rise to an induced emf • This is called a motional emf because it originated from the motion of charges through a magnetic field • If the rod stops moving, the magnetic force vanishes • Then the attractive electric force reunites the positive and negative charges and the emf disappears

EMF Induced in Moving Conductor-4 • If the rod is sliding on conducting rails that form part of a closed circuit • L is the length of the rod between the rails • Due to the EMF, electrons flow in a clockwise direction around the circuit • Positive charge would flow in the direction opposite to the electron flow • The conventional current I is drawn counterclockwise.

Motional EMF • v, B, and L are mutually perpendicular • Eq = (E/L)q = qv. B • E = v. BL

Problem 1 - 1 • The rod is moving at a speed of 5. 0 m/s in a direction perpendicular to a 0. 80 -T magnetic field. • The rod has a length of 1. 6 m • The rod and rails have no resistance • Light bulb has a resistance of 96 ohms

Problem 1 - 2 • (a) Find the EMF produced by the rod • (b) Find the current induced in the circuit • (c) Find the electric power delivered to the bulb

Problem 1 - 3 • (a) The motional EMF is given by ü E = v. BL = (5. 0 m)(0. 80 T)(1. 6 m) = 6. 4 V • (b) Use Ohm’s Law to find the current ü I = E/R = 6. 4 V/96 ohms = 0. 067 A • (c) Electric power is gotten from power formula ü P = IE = (0. 067 A)(6. 4 V) = 0. 43 W

Motional EMF & Electrical Energy-1 • • • Motional EMF arises because a magnetic force acts on the charges in a conductor that is moving through a magnetic field Whenever this EMF causes a current, a second magnetic force enters the picture The second force arises because the current I in the rod is perpendicular to the magnetic field The current and rod experiences a magnetic force F whose magnitude is given by F = ILB The direction of F is specified by RHR-1 and is opposite to the velocity v of the rod

Motional EMF & Electrical Energy-2 • By itself, F would slow down the rod • To keep the rod moving to the right with a constant velocity, a counterbalancing force must be applied to the rod by an external agent, such as the hand • This force is labeled Fhand • What provided the electrical energy that the light bulb used? • The provider is external agent that applies the counterbalancing force needed to keep rod moving

PHY 102: Lecture 7 Electromagnetic Induction 7. 3 Magnetic Flux

Motional EMF and Magnetic Flux - 1 • Motional EMF, can be described in terms of magnetic flux • Magnetic flux brings together the magnetic field and the surface through which it passes

Motional EMF and Magnetic Flux - 2 • E=v. BL • E=

Motional EMF and Magnetic Flux - 3 • The quantity BA is the magnetic flux • E=

Motional EMF and Magnetic Flux - 4 • This is usually written as • E= • The minus sign ensures that the polarity of the induced EMF sends the induced current in the proper direction so as to give rise to this opposing magnetic force

General Expression for Magnetic Flux • is angle between the magnetic field B and the normal to surface • SI Unit for Flux: Weber (Wb)

Problem 2 - 1 • A rectangular coil of wire is in a constant magnetic field whose magnitude is 0. 50 T • The coil has an area of 2. 0 m 2 • Determine the magnetic flux for the = 00, 60. 00, and 90. 00

Problem 2 - 2

Faraday’s Law of Electromagnetic Induction • Average EMF E induced in a coil of N loops is • E= • is the change in magnetic flux through one loop • t is the time interval during which the change occurs • The term / t is the average time rate of change of the flux that passes through one loop • SI Unit of Induced EMF: volt (V)

Problem 3 - 1 • A coil of wire consist of 20 turns each of which has an area of 1. 5 x 10 -3 m 2 • A magnetic field is perpendicular to the surface of each loop at all times § f = i = 0 0 • At time ti = 0 s, the magnitude of the field at the coil is Bi = 0. 050 T • At a later time tf = 0. 10 s, the magnitude of the field at the coil has increased to B = 0. 060 T • (a) Find average emf induced in the coil during this time • (b) What would be the value of the average induced emf if the magnitude of the magnetic field decreased from 0. 060 T to 0. 050 T in 0. 10 s

Problem 3 - 2 • (a) • E=

Problem 3 - 3 • (b) • E=

Problem 4 - 1 • A coil of wire has an area of 0. 020 m 2 and consists of 50 turns • At ti = 0 s the coil is oriented so the normal to its surface is parallel ( i = 00) to a constant magnetic field of magnitude of 0. 18 T • The coil is then rotated through an angle of = 30. 00 in a time of 0. 10 s • (a) Determine the average emf • (b) What would be the induced emf if the coil were returned to its initial orientation in 0. 10 s?

Problem 4 - 2 • (a) • E=

PHY 102: Lecture 7 Electromagnetic Induction 7. 4 Lenz’s Law

Lenz’s Law - 1 • An induced emf drives current around a circuit just as the EMF of a battery does • With an induced emf, conventional current is directed out of the positive terminal, through the attached device, and into the negative terminal • A method is needed for determining the algebraic sign of the induced emf, so the terminals can be identified • It will be helpful to keep in mind that the net magnetic field penetrating a coil of wire results from two contributions • One is the original magnetic field that produces the changing flux that leads to the induced emf • The other arises because of the induced current, which, like any current creates its own magnetic field • Field created by induced current is induced magnetic field

Lenz’s Law - 2 • Lenz’s Law • The induced emf resulting from a changing magnetic flux has a polarity that leads to an induced current whose direction is such that the induced magnetic field opposes the original flux change

Applying Lenz’s Law • Determine whether the magnetic flux that penetrates a coil is increasing or decreasing • Find what the direction of the induced magnetic field must be so that it can oppose the change in flux by adding to or subtracting from the original field • Having found the direction of the induced magnetic field, use RHR-2 to determine the direction of the induced current • The polarity of the induced emf can be assigned because conventional current is directed out of the positive terminal • The induced field is not always opposite to the external field, because Lenz’s law requires only that it must oppose the change in the flux that generates the emf

Problem 5 - 1 • Figure shows a magnet approaching a loop of wire • The external circuit attached to the loop consists of resistance R • Find the direction of the induced current and the polarity of the induced emf

Problem 5 - 2 • Apply Lenz’s law • The change in magnetic flux must be opposed by the induced magnetic field • The magnitude of the magnetic field at the loop is increasing as the magnet approaches • Magnetic flux through the loop is also increasing

Problem 5 - 3 • To oppose the increase in flux, the direction of the induced magnetic field must be opposite to the field of the bar magnet • The field of the bar magnet passes through the loop from left to right • The induced field must pass through the loop from right to left • The induced current must be directed counterclockwise around the loop, when viewed from the side nearest the magnet

Problem 5 - 4 • Remember RHR-1 and RHR-2 are for + charges • The motion of + charges is the conventional current • Loop behaves as a source of emf • Conventional current flows from A to B through resistor • Point A must be the positive terminal

Problem 6 - 1 • There is a constant magnetic field in a rectangular region of space • Field is perpendicular into the page • Outside this region there is no magnetic field • A copper ring slides through the region • For each position, find whether an induced current exists in the ring and its direction

Problem 6 - 2 • Position 1 • No flux passes through the ring because the magnetic field is zero outside the rectangular region • There is no change in flux and no induced emf or current

Problem 6 - 3 • • Position 2 Ring moves into the field region The flux increases There is an induced emf and current • The induced field must point out of the page opposite to the external field • RHR-2 indicates that the direction of the induced current must be counterclockwise

Problem 6 - 4 • Position 3 • The flux passes through the moving ring • There is no induced emf or current because the flux remains constant in the rectangular region • Flux must change as time passes to have induced emf

Problem 6 – 5 a • Position 4 • As ring leaves the field region, the flux decreases • The induced field must oppose this change • The induced field must point in the same direction as the external field – into the paper • With this orientation the induced field will increase the net magnetic field through the ring and increase the flux

Problem 6 – 5 b • Position 4 • To produce an induced field pointing into the paper, RHR-2 indicates that the induced current must be clockwise • Position 5 • There is no induced current

PHY 102: Lecture 7 Electromagnetic Induction 7. 5 Mutual and Self Inductance

Mutual Inductance - 1 • Two coils of wire are placed close to each other • The primary coil and the secondary coil • The primary coil is connected to an ac generator, which sends an alternating current Ip through it • The secondary coil is not attached to a generator

Mutual Inductance - 2 • Current-carrying primary coil is an electromagnet • It creates a magnetic field in the surrounding region • If the two coils are close to each other, a significant fraction of this magnetic field penetrates the secondary coil and produces a magnetic flux • The flux is changing, since the current in the primary coil and its associated magnetic field are changing • Because of the change in flux, an emf is induced in the secondary coil • The effect in which a changing current in one circuit induces an emf in another circuit is called mutual induction

Mutual Inductance - 3 • Faraday’s law of electromagnetic induction • Average emf Es in the secondary coil is proportional to the change in flux s passing through it • Ns s = MIp or M = Ns s/Ip • M is mutual inductance. • Es= • SI Unit of M: henry (H)

Self-Inductance - 1 • An emf can be induced in a current-carrying coil by a change in the magnetic field that the current itself produces • The alternating current creates an alternating magnetic field that, in turn, creates a changing flux through the coil • The change in flux induces an emf in the coil • The effect in which a changing current in a circuit induces an emf in the same circuit is referred to as self-induction

Self-Inductance - 2 • N = LI • L = N /I • L is the self inductance • E=