Electricity and Magnetism Chapter 23

Hans Christian Oersted 1777 – 1851 Best known for observing that a compass needle deflects when placed near a wire carrying a current First evidence of a connection between electric and magnetic phenomena

Magnetic Fields – Long Straight Wire A current-carrying wire produces a magnetic field The compass needle deflects in directions tangent to the circle The compass needle points in the direction of the magnetic field produced by the current

Direction of the Field of a Long Straight Wire Right Hand Rule #1 Grasp the wire in your right hand Point your thumb in the direction of the current Your fingers will curl in the direction of the field

Magnitude of the Field of a Long Straight Wire A straight current-carrying wire generates a cylindrical magnetic field in the space surrounding it. The magnitude of the field at a distance r from a wire carrying a current of I is µo = 4  x 10-7 T.m / A µo is called the permeability of free space The magnetic field of a long straight wire in a vacuum is

Example 1 The overhead power cable for a street trolley is strong horizontally 10 m above the ground. A long straight section of it carries 100 amps dc due west. Describe the magnetic field produced by the current, and determine its value at ground level just under the wire. Compare that to the strength of the Earth's field. Given: r = 10 m and I = 100 A Find: B

Example 1 Solution: We've got a straight current-carrying wire and that produces a known B-field. See diagram on last slide with the current assumed heading west. At ground level (at a point beneath the westerly current), the Right-Hand-Current Rule tells us that B points due south. We already have an expression for the field of a long straight current-carrying wire in terms of I and r. m0 = 4p x 10-7 T-m/A. Which is only 4% of the Earth's field.

Force Between Two Conductors The magnetic fields created by moving charges can interact and create forces much like the electric forces between those charges. Parallel conductors carrying currents in the same direction attract each other Parallel conductors carrying currents in the opposite directions repel each other

Magnetic Force Between Two Parallel Conductors The force on wire 1 is due to the current in wire 1 and the magnetic field produced by wire 2 The force per unit length is:

Example 2 What is the force per unit length experienced by each of two extremely long parallel wires carrying equal 1.0-A currents in opposite directions while separated by a distance of 1 m in vacuum? Given: I1 = I2 = 1.0 A and d = 1 m Find: FM

Example 2 Make a drawing. The defining equation is FM/l = 2 x 10-7N/m, repulsion

André-Marie Ampère 1775 – 1836 Credited with the discovery of electromagnetism Relationship between electric currents and magnetic fields Mathematical genius evident by age 12

Ampère’s Law ***THE MORE CURRENT….THE STRONGER THE MAGNETIC FIELD*** André-Marie Ampère found a procedure for deriving the relationship between the current in an arbitrarily shaped wire and the magnetic field produced by the wire Ampère’s Circuital Law Sum over the closed path ***THE MORE CURRENT….THE STRONGER THE MAGNETIC FIELD***

Magnetic Field of a Current Loop The strength of a magnetic field produced by a wire can be enhanced by forming the wire into a loop All the segments, ∆x, contribute to the field, increasing its strength

Magnetic Field of a Current Loop – Total Field

Magnetic Field of a Current Loop – Equation The magnitude of the magnetic field at the center of a circular loop with a radius R and carrying current I is With N loops in the coil, this becomes

Magnetic Field of a Solenoid If a long straight wire is bent into a coil of several closely spaced loops, the resulting device is called a solenoid It is also known as an electromagnet since it acts like a magnet only when it carries a current

Magnetic Field of a Solenoid, 2 The field lines inside the solenoid are nearly parallel, uniformly spaced, and close together This indicates that the field inside the solenoid is nearly uniform and strong The exterior field is nonuniform, much weaker, and in the opposite direction to the field inside the solenoid

Magnetic Field in a Solenoid, 3 The field lines of the solenoid resemble those of a bar magnet

Example 3 A 20-cm-long solenoid with a 2.0-cm inside diameter is tightly wound on a hollow quartz cylinder. There are several layers with a total of 20 x 103 turns per meter of a niobium-tin wire. The device is cooled below its critical temperature and becomes superconducting. Since the wire is then without resistance, it can easily carry 30 A and not develop any I2R losses. Compute the approximate field inside the solenoid near the middle. What is its value at either end? Given: n = 20 x 103 m-1, I = 30 A, and D = 2.0 cm Find: Bz

Bz ≈ m0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A) Example 3 Solution: We've got a current-carrying solenoid and that produces a known B-field. The solenoid is long and narrow and will obey the approximations that led to the last equation Using m0 = 1-257 X 10-6 T-m/A, Bz ≈ m0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A) Bz ≈ 0.75 T Which is a formidable field, over 104 times that of the Earth. The field at either end is about half this, 0.38 T

Moving Charges and Magnetism Moving charge creates B The orientation (direction) of B depends upon the orientation (direction) of I WITHIN atoms, electrons move in different (often opposing) directions, thus individual B usually cancel out BETWEEN atoms that DO have B fields, those FIELDS may oppose each other and cancel out **Only in materials that have BOTH alignments do we see magnetic properties*** FERROMAGNETIC

Magnetic Force

Right Hand Rule #2 Place your fingers in the direction of Curl the fingers in the direction of the magnetic field, Your thumb points in the direction of the force, , on a positive charge If the charge is negative, the force is opposite that determined by the right hand rule Maximum force is

Finding the Direction of Magnetic Force Experiments show that the direction of the magnetic force is always perpendicular to both and which form a plane. Fmax occurs when is perpendicular to F = 0 when is parallel to

Example 4 A conventional water-cooled electromagnet produces a 3.0-T uniform magnetic field in the 4-in. gap between its flat pole pieces. The field is aligned horizontally pointing due north. A proton is fired into the field region at a speed of 5.0 x 106 m/s. It enters traveling in a vertical north-south plane, heading north and downward at 30° below the horizontal. Compute the force vector acting on the proton at the moment it enters the field. Given: A proton with v = 5.0 X 106 m/s, at 30° below the horizontal in the northerly direction, and B = 3.0 T, north Find: FM

FM = (+1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T)(sin 30°) Example 4 Solution: Here a charged particle is moving in a B-field and that should call to mind v x B. First, make a drawing The proton has a positive charge of +1.60 X 10-19 C and so v x B is due east, FM is due east The basic force-on-a-moving-charge relationship is FM = qvB sin q The angle between v and B is q = 30° and so with q = qe FM = qe vB sin q FM = (+1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T)(sin 30°) Fm= 1.2 x 10-12 N

Force on a Charged Particle in a Magnetic Field Consider a particle moving in an external magnetic field so that its velocity is perpendicular to the field The force is always directed toward the center of the circular path The magnetic force causes a centripetal acceleration, changing the direction of the velocity of the particle

Force on a Charged Particle Charged particle entering perpendicular to uniform magnetic field and experiences centripetal acceleration Equating the magnetic and centripetal forces: Solving for R: R is proportional to the momentum of the particle and inversely proportional to the magnetic field Sometimes called the cyclotron equation

Particle Moving in an External Magnetic Field If the particle’s velocity is not perpendicular to the field, the path followed by the particle is a spiral The spiral path is called a helix

Electromagnets and the Electric Motor 23.2

Electromagnets Electromagnets are created by an electric current travelling through a solenoid. As such, their strength and direction can be controlled. Strength can be increased by: Increasing current (I) Increasing the number of coils (N) – although this means a longer total wire length and thus more resistance. Increasing the core material’s permeability.

Electromagnets

Magnetic Force on Current Carrying Conductor A force is exerted on a current-carrying wire placed in a magnetic field The current is a collection of many charged particles in motion The direction of the force is given by right hand rule #2

Force on a Wire The blue x’s indicate the magnetic field is directed into the page The x represents the tail of the arrow Blue dots would be used to represent the field directed out of the page The • represents the head of the arrow In this case, there is no current, so there is no force

Force on a Wire B is into the page The current is up the page The force is to the left

Force on a Wire B is into the page The current is down the page The force is to the right

Force on a Wire, equation The magnetic force is exerted on each moving charge in the wire The total force is the sum of all the magnetic forces on all the individual charges producing the current FM = B I l sin q q is the angle between and the direction of I The direction is found by the right hand rule, placing your fingers in the direction of I instead of

Example 5 A flat, horizontal rectangular loop of wire is positioned, as shown in Fig. 19.32a, in a 0.10-T uniform vertical magnetic field. The sides of the rectangle are F-C equal to 30 cm and C-­ equal to 20 cm. Determine the total force acting on the loop when it carries a current of 1.0 A. Given: B = 0.10 T, F-C = 30 cm, C-D = 20 cm, and I = 1.0 A Find: FM.

Example 5 Solution: We've got a current-carrying loop in a S-field – FM = IlB sin q First, make a drawing Current travels from the positive terminal of the battery clockwise around the circuit. The directions of the forces on each segment are arrived at via v x B and are indicated in the diagram. Because the forces on segments F-C and D-E are equal and opposite, they cancel. The total force acting on the loop is the force acting on segment C-D

FM = IlB sin q = (1.0 A) (0.20m) (0.10 T) (sin 90°) FM = 0.020 N Example 5 FM = IlB sin q = (1.0 A) (0.20m) (0.10 T) (sin 90°) FM = 0.020 N

Torque on a Current Loop f Applies to any shape loop N is the number of turns in the coil Torque has a maximum value of NIAB When f = 90° Torque is zero when the field is parallel to the plane of the loop

Electric Motor An electric motor converts electrical energy to mechanical energy The mechanical energy is in the form of rotational kinetic energy An electric motor consists of a rigid current-carrying loop that rotates when placed in a magnetic field

Electric Motor, 2 The torque acting on the loop will tend to rotate the loop to smaller values of q until the torque becomes 0 at q = 0° If the loop turns past this point and the current remains in the same direction, the torque reverses and turns the loop in the opposite direction

Electric Motor, 3 To provide continuous rotation in one direction, the current in the loop must periodically reverse In ac motors, this reversal naturally occurs In dc motors, a split-ring commutator and brushes are used Actual motors would contain many current loops and commutators

Electric Motor

Electric Motor, final Just as the loop becomes perpendicular to the magnetic field and the torque becomes zero, inertia carries the loop forward and the brushes cross the gaps in the ring, causing the current loop to reverse its direction This provides more torque to continue the rotation The process repeats itself ***This is obviously much easier if the CURRENT itself periodically reverses – AC electricity instead of DC***

Induction and the Electric Generator 23.3

Electromagnetic Induction Moving charge(s) creates (induces) a magnetic field (B) A moving (changing) magnetic field (B) creates (induces) a current (I) Whether B is increasing or decreasing near a wire determines the current’s direction.

Michael Faraday 1791 – 1867 Great experimental scientist Invented electric motor, generator and transformers Discovered electromagnetic induction Discovered laws of electrolysis

Faraday’s Experiment – Set Up A current can be produced by a changing magnetic field First shown in an experiment by Michael Faraday A primary coil is connected to a battery A secondary coil is connected to an ammeter

Faraday’s Experiment The purpose of the secondary circuit is to detect current that might be produced by the magnetic field When the switch is closed, the ammeter reads a current and then returns to zero When the switch is opened, the ammeter reads a current in the opposite direction and then returns to zero When there is a steady current in the primary circuit, the ammeter reads zero

Faraday’s Conclusions An electrical current is produced by a changing magnetic field The secondary circuit acts as if a source of emf were connected to it for a short time It is customary to say that an induced emf is produced in the secondary circuit by the changing magnetic field

Magnetic Flux (Ф) The emf is actually induced by a change in the quantity called the magnetic flux rather than simply by a change in the magnetic field Magnetic flux (Ф) is defined in a manner similar to that of electrical flux – the density of the magnetic field lines in a given area ofspace Magnetic flux is proportional to both the strength of the magnetic field passing through the plane of a loop of wire and the area of the loop

Magnetic Flux, 2 You are given a loop of wire The wire is in a uniform magnetic field The loop has an area A The flux is defined as ΦM = BA = B A cos θ θ is the angle between B and the normal to the plane

Magnetic Flux, 3 When the field is perpendicular to the plane of the loop, as in a, θ = 0 and ΦB = ΦB, max = BA When the field is parallel to the plane of the loop, as in b, θ = 90° and ΦB = 0 The flux can be negative, for example if θ = 180° SI units of flux are T. m² = Wb (Weber)

Magnetic Flux, final The flux can be visualized with respect to magnetic field lines The value of the magnetic flux is proportional to the total number of lines passing through the loop When the area is perpendicular to the lines, the maximum number of lines pass through the area and the flux is a maximum When the area is parallel to the lines, no lines pass through the area and the flux is 0

Electromagnetic Induction – An Experiment When a magnet moves toward a loop of wire, the ammeter shows the presence of a current (a) When the magnet is held stationary, there is no current (b) When the magnet moves away from the loop, the ammeter shows a current in the opposite direction (c) If the loop is moved instead of the magnet, a current is also detected

Electromagnetic Induction – Results of Experiment A current is set up in the circuit as long as there is relative motion between the magnet and the loop (changing magnetic field) The same experimental results are found whether the loop moves or the magnet moves The current is called an induced current because is it produced by an induced emf

Faraday’s Law and Electromagnetic Induction The average induced emf for a one turn loop A single loop of wire will experience an induced voltage that equal the time rate of change of magnetic flux through it at any given instant. For N turns of wire, the average induced emf is Which is known as Faraday’s Induction Law Eav Eav

Example 6 A circular flat coil of 200 turns of wire encloses an area of 100 cm2. The coil is immersed in a uniform perpendicular magnetic field of 0.50 T that penetrates the entire area. If the field is shut off so that it drops to zero in 200 ms, what is the average induced emf? Given that the coil has a resistance of 25 W, what current will be induced in it? Given: N = 200, A = 100 cm2, Bi = 0.50 T, Bf = 0, ∆t = 200 ms, and R = 25 W Find: The induced emf and current I

Example 6 Solution: Faraday's Law problem The time rate-of-change of the flux equals the emf. The B-field links the entire area perpendicularly, the initial flux is FM = BA = (0.50T)(0.0100 m2) = 0.0050 T-m2 and so ∆FM = - 0.005 0 T- m2 since the final flux is zero

Example 6 Thus, From Ohm's Law: E

Applications of Faraday’s Law Electric Guitar A vibrating string induces an emf in a coil A permanent magnet inside the coil magnetizes a portion of the string nearest the coil As the string vibrates at some frequency, its magnetized segment produces a changing flux through the pickup coil The changing flux produces an induced emf that is fed to an amplifier

Applications of Faraday’s Law Apnea Monitor The coil of wire attached to the chest carries an alternating current An induced emf produced by the varying field passes through a pick up coil When breathing stops, the pattern of induced voltages stabilizes and external monitors sound an alert

Application of Faraday’s Law Motional emf A straight conductor of length ℓ moves perpendicularly with constant velocity through a uniform field The electrons in the conductor experience a magnetic force FM = qvB sin q = qvB The electrons tend to move to the lower end of the conductor

Motional emf As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor As a result of this charge separation, an electric field is produced in the conductor Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force There is a potential difference between the upper and lower ends of the conductor

Motional emf, cont The potential difference between the ends of the conductor can be found by E = v B l The upper end is at a higher potential than the lower end A potential difference is maintained across the conductor as long as there is motion through the field If the motion is reversed, the polarity of the potential difference is also reversed Since emf = E l, the electric field in the wire, which exactly counters the motional emf is E = v B

Example 7 A 1.0-meter-long wire held in a horizontal east-west orientation is dropped at a place where the Earth's magnetic field is 2.0 x 10 -5 T, due north. B = 2.0 x 10-5 T 1.0 m N S E W Determine the induced emf 4.0 s after release. Given: l = 1.0 m, t = 4.0 s. and B = 2.0 X 10 -5 T Find: The resulting emf

Example 7 Solution: Because a wire is moving through a magnetic field and we want the emf, this problem involves the expression E = vBl. We have B and land need v The speed at t = 4.0 s v = v0 + gt = 0 + (9.81 m/s2)(4.0 s) = 39.2m/s E = vBl = (39.2 m/s)(2.0 x 10 -5T)(l.0m) = 0.78 mV

Motional emf in a Circuit Assume the moving bar has zero resistance As the bar is pulled to the right with a given velocity under the influence of an applied force, the free charges experience a magnetic force along the length of the bar This force sets up an induced current because the charges are free to move in the closed path

Motional emf in a Circuit The changing magnetic flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop The induced, motional emf, acts like a battery in the circuit E = v B l and

Generators Alternating Current (AC) generator Converts mechanical energy to electrical energy Consists of a wire loop rotated by some external means – a Turbine There are a variety of sources that can supply the energy to rotate the loop These may include falling water, heat by burning coal to produce steam

AC Generators Basic operation of the generator As the loop rotates, the magnetic flux through it changes with time This induces an emf and a current in the external circuit The ends of the loop are connected to slip rings that rotate with the loop Connections to the external circuit are made by stationary brushes in contact with the slip rings

AC Generators, final E Emax The emf generated by the rotating loop can be found by E = 2 N B l v sin θ If the loop rotates with a constant angular speed, ω, and N turns E = N A B ω sin ω t E = E max when loop is parallel to the field E = 0 when when the loop is perpendicular to the field Emax

AC Generators – Detail of Rotating Loop The magnetic force on the charges in the wires AB and CD is perpendicular to the length of the wires An emf is generated in wires BC and AD The emf produced in each of these wires is E= B l v= B l sin θ

AC Generators Can also consist of a rotating turbine which passes constantly-reversing (and changing) magnetic fields past an electromagnet. This changes the electromagnet’s magnetic field and induces AC current in its coil

DC Generators Components are essentially the same as that of an ac generator The major difference is the contacts to the rotating loop are made by a split ring, or commutator

DC Generators E The output voltage always has the same polarity The current is a pulsing current To produce a steady current, many loops and commutators around the axis of rotation are used The multiple outputs are superimposed and the output is almost free of fluctuations E

Example 8 A simple single-coil dc generator rotates at a constant frequency of 60 Hz in a 0.40-T magnetic field. Given that the coil has 10 turns and encompasses an area of 1200 cm2, what will be its maximum emf? Given: f = 60 Hz, A = 1200 cm2, B = 0.40 T, and N = 10 Find: Em.

Em = (10)(1200X 10-4m2)(0.40T)(376.99 rad/s) = 0.18 kV Example 8 A coil is rotating in a B-field with a specified frequency – sinusoidal emf The basic formula for the emf of an ac generator is Eq. (20.7) The maximum emf (Em) is the amplitude of the oscillating voltage given by E = NAB w sinwt namely E m= NAB w w = 2pf= 2p(60 Hz) = 376.99 rad/s Em = (10)(1200X 10-4m2)(0.40T)(376.99 rad/s) = 0.18 kV

Motors Motors are devices that convert electrical energy into mechanical energy A motor is a generator run in reverse A motor can perform useful mechanical work when a shaft connected to its rotating coil is attached to some external device

Transformers Use EM Induction to control voltages by “Stepping-Up” or “Stepping-Down” the voltage. Using different numbers of coils of wire wrapped around a common core, the induced emf in one part of the coil is transferred to the other coil. Larger-N coil has more voltage than the… Smaller-N coil (for the same B-Field generated in the core V2/V1 = N2/N1

Transformers