a) head-on view b) side view c) side view Use the right hand rule for forces to confirm the direction of the force in each case.

Electromagnetism Magnetic Force on a Wire Magnetic Field around a Bar Magnet Direction of magnetic field lines: the direction that the North pole of a small test compass would point if placed in the field (N to S) If a wire with current flowing through it is placed in an external magnetic field, it will experience a force. Why? Two magnetic fields around wire and from external magnet will either attract or repel What is the cause of magnetic fields? Moving electric charges Therefore: current in a wire will produce a magnetic field The Right Hand Rule for the Magnetic Force on a Current-Carrying Conductor in a Magnetic Field Flat Hand: thumb and fingers at right angles Fingers: external B field north to south Thumb: current The Right Hand Rule for the Magnetic Field around a Wire Thumb: direction of conventional current Fingertips: direction of magnetic field tangent to circle Palm: force palm pushes Maximum force occurs when: current is perpendicular to B field No force occurs when: current is parallel to B field a) head-on view b) side view c) side view Use the right hand rule for forces to confirm the direction of the force in each case. Draw concentric circles with increasing spacing and arrows in circular fashion. Draw concentric circles around wire. Draw dots and crosses Alternate Right Hand Rule for Loops Fingertips: direction of current Component fields Resultant field Your turn Magnitude of the magnetic force on a wire: F = BIL sin θ Where θ is angle between current and B field Magnetic field strength Magnetic field intensity Magnetic flux density Units: Tesla (T) Thumb: points North Note that a wire loop acts like a: bar magnet Definition of magnitude of magnetic field (#1): B = F / (IL sin θ) The ratio of the magnetic force on a wire to the product of the current in the wire, the length of the wire and the sine of the angle between the current and the magnetic field Solenoid: coil of wire many loops Note that a solenoid acts like a: bar magnet Draw the magnetic field around this solenoid. Use alternate RHR to find North. 1 Find the magnitude and direction of the force on the wire segment confined to the gap between the two magnets as shown when the switch is closed. The strength of the magnetic field in the gap is 1.9 T. 0.62 N up 2

Magnetic Force on a Moving Charged Particle Motion of a Charged Particle in a Magnetic Field Why is there a magnetic force on a charged particle as it moves through a magnetic field? Moving charged particle creates its own magnetic field two magnetic fields interact 1. A charged particle will follow a circular path in a magnetic field since the magnetic force is always perpendicular to the velocity. The Hand Rule for the Magnetic Force on a Charge moving in a Magnetic Field Flat Hand: thumb and fingers at right angles Fingers: external B field north to south 2. The magnetic force does no work on the particle since the magnetic force is always perpendicular to the motion. Right Hand: positive charge Left Hand: negative charge Thumb: velocity Palm: force palm pushes 3. The particle accelerates centripetally but maintains a constant speed since the magnetic force does no work on it. Maximum force occurs when: velocity is perpendicular to B field No force occurs when: velocity is parallel to B field Radius of Circular Path a) Sketch the paths of a slow and a fast moving proton at constant speed. b) Sketch the path of a proton that is slowing down and one that is speeding up. Find the direction of the magnetic force on each particle below as each enters the magnetic field shown. Σ F = m a c F B = m v 2 /r Q v B = m v 2 /r r = mv/ qb Magnitude of the magnetic force on a moving charged particle: Definition of magnitude of magnetic field (#2): B = F / (qv sin θ) F = qvb sin θ Where θ is angle between v and B The ratio of the force on a charged particle moving in a magnetic field to the product of the particle s charge, velocity and sine of the angle between the direction of the magnetic field and velocity. Electric Field c) How would the radius of the path change if the particle were an alpha particle? Comparing Electric and Magnetic Fields and Forces Magnetic Field A proton in a particle accelerator has a speed of 5.0 10 6 m/s. The proton encounters a magnetic field whose magnitude is 0.40T and whose direction makes an angle of θ = 30.0 with respect to the proton's velocity. Find the magnitude of the magnetic force on the proton and the proton s acceleration. How would these change if the particle was an electron? a) 1.6 x 10-13 N b) 9.6 x 10 13 m/s 2 c) 1.8 x 10 17 m/s 2, same force but opposite direction 3 draw paths for stationary, parallel and perp charges 4

Electric Fields and Magnetic Fields The Mass Spectrometer 1. A proton is released from rest near the positive plate and leaves through a small hole in the negative plate where it enters a region of constant magnetic field of magnitude 0.10T. The electric potential difference between the plates is 2100 V. a) Describe the motion of the proton while in the electric field constant acceleration in a straight line b) Describe the motion of the proton while in the magnetic field constant acceleration and constant speed circular path A mass spectrometer is a device used to measure the masses of isotopes. Isotopes of the same element have the same charge and chemical properties so they cannot be separated by using chemical reactions but have different masses and so can be separated by a magnetic field. A common type of mass spectrometer is known as the Bainbridge mass spectrometer and its main parts are shown below. Ion Source: source of charged isotopes same charge different mass c) Find the speed of the proton as it enters the magnetic field. Use conserv. Of energy a) 6.3 x 10 5 m/s Velocity selector: so all ions have the same speed Magnetic deflection chamber: radius is proportional to mass d) Find the radius of the circular path of the proton in the magnetic field. b) 6.6 x 10-2 m 1. A singly charged ion with mass 2.18 x 10-26 kg moves without deflection through a region of crossed magnetic and electric fields then is injected into a region containing only a magnetic field, as shown in the diagram, where it is deflected until it hits a photographic plate. The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both regions is 0.930 T. Determine the sign of the charge and calculate where the ion lands on the photographic plate. 2. A Velocity Selector is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance. Sign: could be either in velocity selector Only positive in deflection chamber velocity selector v = E/B v = 1.0 x 10 3 m/s magnetic chamber r = mv/qb r = 1.5 x 10-4 m a) Determine the magnitude and direction of an electric field that will apply and electric force to balance the magnetic force on the proton. d = 3.0 x 10-4 m perp to v and B from bottom to top of page ΣF = 0 F B F e = 0 F B = F e qvb = Eq E = vb 2. A hydrogen atom and a deuterium atom (an isotope of hydrogen) move out of the velocity selector and into the region of a constant 0.10 T magnetic field at point S, as shown below. Each has a speed of 1.0 x 10 6 m/s. Calculate where they each hit the photographic plate at P. Hydrogen = 0.20 m Deuterium = 0.41 m b) What is the resulting speed and trajectory of the proton? v = E/B in a straight line 5 6

Electromagnetic Induction Two Opposing Forces I In 1819, Hans Christian Oersted discovered that a magnetic compass experiences a force in the vicinity of an electric current that is, that electric currents produce magnetic fields. Because nature is often symmetric, this led many scientists to believe that magnetic fields could also produce electric currents, a concept known as electromagnetic induction. I v F B The magnetic force acts to oppose the applied force, like drag or friction. Why does moving a wire through a magnetic field induce a current in the wire? Free electrons in the wire are charged particles moving through a magnetic field so there is a qvb force on them causing them to move resulting in a current. Derivation of formula for EMF induced in a moving wire A straight conductor is moved at constant velocity perpendicular to a uniform magnetic field. Palm pushes current up F B = qvb An applied force (F app) in the direction of the velocity induces an emf which causes current to be pushed upwards. Palm pushes bar back F B = BIl The induced current now generates a magnetic field around the moving bar that causes a magnetic force (F B) on itself. At a constant speed, F app = F B = BIl 1. Electrons in the moving conductor experience a downward magnetic force and migrate to the lower end of the conductor, leaving a net positive charge at the upper end. 2. As a result of this charge separation, an electric field is built up in the conductor. F B = qvb LHR for electrons to show direction of force E Suppose a rod is moving at a constant speed of 5.0 m/s in a direction perpendicular to a 0.80-T magnetic field as shown. The rod has a length of 1.6m and negligible electrical resistance. The rails also have negligible resistance. The light bulb, however, has a resistance of 96 Ω. Find: a) the emf produced by the motion of the rod 6.4 V 3. Charge builds up until the downward magnetic force is balanced by the upward electric force due to the electric field. At this point, the charges stop flowing and are in equilibrium. 4. Because of this charge separation, a potential difference is set up across the conductor. F B = F e qvb = Eq E = vb V = E d = E l V= vb l ε= B l v (b) the magnitude and direction of the induced current in the circuit 0.067 A CCW e) How much external force is applied to keep the rod moving at this constant speed? 0.086 N If the conductor is connected to a complete circuit, the induced emf will produce an induced current. is equivalent to Amount of Current The amount of induced current in the circuit is given by ε= B l v ε = I R Direction of Current The direction of the induced emf and induced current can be found from the right hand rule for forces to find the force on a positive charge in the conductor. c) the electrical power delivered to the bulb 0.43 W d) the energy used by the bulb in 60.0 s. 26 J f) How much work is done by the applied force in 60.0 seconds? 26 J g) What happens to this work? Converted to electrical energy I = Blv/R 7 8

Magnetic Flux EMF Induced by a Time-Changing Flux Magnetic Flux Number of field lines Symbol: Φ Units: Weber (Wb) = T m 2 Magnetic Flux Density (field strength/intensity) number of field lines per unit area Formula: B = Φ/ A or Φ = B A Symbol: B Units: Wb/m 2 = T (tesla) Angle Dependence of Flux: What is the amount of magnetic flux if the field lines are not perpendicular to the cross-sectional area? Only the perpendicular component of the magnetic field contributes to the magnetic flux. Normal line: line perpendicular to plane of cross-sectional area Moving a magnet towards a coil will increase the magnetic flux linking the coil and will induce an emf and a current in a certain direction. Holding the magnet stationary will not change the amount of magnetic flux linking the coil and so will not induce an emf or current. Methods of inducing an EMF by a time-changing flux Moving the magnet away from the coil will decrease the magnetic flux linking the coil and will induce an emf and a current in the opposite direction. Formula: Φ = (B cos θ) A = B A cos θ 2. Move magnet or coil 2. Rotate coil 1. Vary magnetic field Angle: θ = angle between normal line and field lines Magnetic Flux(Φ) - product of the magnetic field strength and a cross-sectional area and the cosine of the angle between the magnetic field and the normal to the area Formula: Φ = B A cos θ Units: T m 2 Magnetic flux linkage (magnetic flux linking a coil): product of magnetic flux through a coil of wire and the number of turns of the wire Faraday s Law: an induced emf is proportional to the rate of change of the flux linkage Formula: ε = - N ( Φ/ t) 1. A coil of area 0.030 m 2 with 300 turns of wire rotates as shown in 0.10 second in a magnetic field of constant 0.25 T strength. a) What is the magnitude of the induced emf? 11.3 V Formula: N Φ = N B A cos θ Units: T m 2 1. A single loop of wire whose cross-sectional area is 0.50 m 2 is located in a 0.20 T magnetic field as shown. Calculate the flux through the loop in each case. a) 0.10 T m 2 b) 0.050 T m 2 b) What is the magnitude of the induced emf if the coil were stationary at 0 0 but the field strength changed from 0.25 T to 0.60 T in 0.10 second? 22.5 V c) 0 2. If the coil of wire in the above example consisted of 50 turns of the wire, calculate the amount of flux linking the coil in each case. a) 5.0 Tm 2 b) 2.5 T m 2 c) 0 9 10

2. A 50 turn coil of wire of area 0.20 m 2 is perpendicular to a magnetic field that varies with time as shown by the graph. 3. If the current in the wire is increasing, in which direction will there be an induced current in the rectangular wire loop? 4. If the wire loop moves away from a steady current in the straight wire, in which direction will there be an induced current in the loop? a) Determine the emf induced in the coil during each time interval. 3 V, 0 V, -1.5 V 5. A conducting loop moves at a constant speed into and through a uniform magnetic field as shown in the diagram. Indicate the direction of the induced current. Graph the flux through the loop and the induced emf as a function of time. b) Sketch a graph of the induced emf vs. time. Emf = - derivative of flux Lenz s Law Finding the Direction of the Induced emf Lenz s Law - The direction of an induced emf is such that it produces a magnetic field whose flux opposes the flux change that induced it. (An emf will be induced so as to keep the net flux constant.) a) Original flux change an increasing flux induces an emf and current. b) Induced flux opposes increasing flux by pointing in opposite direction thus current is in direction shown. c) Result - two magnetic fields acting to keep net flux constant. 6. If a clockwise current through the primary coil is increasing with time, what effect will this have on the secondary coil? 7. Determine the direction of the current in the solenoid in each case. 8. Determine the direction(s) of the induced current as the magnet falls through the loop. 1. If the magnetic field linking this coil is decreasing with time, in which direction is the induced current? 2. The diagrams show a conducting ring that is placed in a uniform magnetic field. Deduce the direction of the induced current in each case if there is (a) an increasing B field (b) a decreasing B field 11 12

Alternating Current Generators Alternating Current Basic Operation: 1. coil of wire is turned by mechanical means in an external magnetic field 2. emf and current are induced in coil as coil cut flux lines 3. current varies in magnitude and direction as flux linkage changes current and emf variations are sinusoidal 4. brushes and rings maintain contact with external circuit without getting tangled Rotation of a Coil in a Uniform Magnetic Field induces an EMF The output of an AC generator is an emf that varies sinusoidally with time. The power output of an AC generator V 0 = peak/ maximum voltage I 0 = peak/ maximum current V = V o sin ωt (where ω = 2πf) I = V/R so I = (V o/r) sin ωt = I o sin ωt P = I V P = (I o sin ωt)(v o sin ωt) P = I o V o sin 2 ωt As the coil rotates, the flux linking it changes Position 1 Maximum EMF and Current 1. sides of coil cut field lines perpendicularly Maximum Power Average Power RMS Values 2. plane of coil is parallel to field lines 3. normal to coil is perpendicular to field lines (90 0 ) P max = I o V o P max = 2 P av P av = ½ I o V o = (I o/ rad 2)(V o / rad2) = I rms V rms I rms = I o / rad 2 V rms = V o / rad 2 Position 2 Minimum EMF and Current 1. sides of coil do not cut field lines perpendicularly (move parallel to them) 2. plane of coil is perpendicular to field lines Root-Mean-Squared values (RMS): The rms value of an alternating current (or voltage) is that value of the direct current (or voltage) that dissipates power in a resistor at the same rate. Mark when the coil is in positions 1 and 2. Sketch the graph of the induced current. 3. normal to coil is parallel to field lines (0 0 ) Sketch a graph of the induced emf for a coil with: twice the frequency of rotation. half the frequency of rotation. 1. In the USA, most household voltage is stated as 120 V at 60 Hz. This is the root-mean-square voltage and the frequency of the AC voltage. Calculate the maximum voltage and mark V o, V rms, on the graph. V o = 170 V 2. In Europe, the mains electricity is rated at 230 V. What is the peak household voltage in Europe? 13 14

The Transformer Rating: rms values are given as the AC values to be used in calculations, as if they were DC values Formula: R = V 0/I 0 = V rms/i rms 1. A stereo receiver applies an AC voltage of 34 V to a speaker. The speaker behaves approximately as if it has a resistance of 8.0 Ω, as the circuit figure indicates. Determine a) the maximum voltage, b) the rms current, According to Michael Faraday s original experiment that first produced electromagnetic induction, an emf and current were only induced in the secondary coil when the switch in the primary coil was being opened or closed, that is, when the current in the primary coil was changing (increasing or decreasing). No emf or current was induced in the secondary coil while the switch was stationary in the open or closed position, that is, when the current was steady or off. Therefore, emf can only be induced in the secondary coil when the magnetic field from the current in the primary coil is building up or dying down, that is, while the magnetic flux is changing. Transformer: a device that increases or decreases AC voltage. Structure and operation of a transformer Your Turn c) the average power for this circuit. a) 48 V b) 4.25 A c) 145 W 2. A 100 W light bulb is designed to operate from a 120 VAC mains. Determine: a) the maximum power of the light bulb 3. A maximum alternating voltage of 170 V is applied across a 50 Ω resistor. Determine: a) the maximum current through the resistor 1. An alternating potential difference (V P) applied across the primary coil creates an alternating current in the primary coil. 2. This creates an alternating magnetic field (time-changing flux) in the primary coil. 3. The soft iron core concentrates the magnetic flux from the primary coil and links it with the secondary coil. 4. The time-changing flux in the secondary coil induces a secondary alternating emf (V S). b) the maximum current drawn by the bulb b) the average power dissipated by the resistor Transformer formula ε P = V P = -N P ( Φ/ t) Step-Up Transformer: If N S > N P, then V S > V P and voltage increases from primary to secondary ε S = V S = -N S ( Φ/ t) a) 200 W b) 1.2 A a) 3.4 A b) 289 W 15 since flux changes are identical V P / V S = N P / N S Voltage and turns in same ratio Step-Down Transformer: If N S < N P, then V S < V P and voltage decreases from primary to secondary 16

How can the voltage increase or decrease without violating the conservation of energy principle? The power input at the primary equals the power output at the secondary. (This assumes 100% efficiency and such a transformer is termed an ideal transformer.) 2. The figure shows a step-down transformer used to light a filament lamp with a resistance of 4.0 Ω under operating conditions. The secondary coil has an effective resistance of 0.2 Ω and the primary current is 150 ma. Calculate: a) the reading on the voltmeter with switch S open 12 V P P = P S Ideal Transformer Formula d) the power taken from the mains supply 36 W V P I P = V S I S V P/ V S = I S / I P Voltage and current in inverse ratio 1. A 120 VAC wall outlet is used to run a small electronic appliance with a resistance of 2.0 Ω, as shown in the diagram. a) Is the transformer a step-up or step-down transformer? Cite evidence for your answer. b) the current in the secondary coil with switch S closed 2.86 A c) the power dissipated in the lamp and the secondary coil 3.27 W and 1.6 W e) the efficiency of the transformer 95% b) How much voltage does the device need? c) If the current in the primary coil is 150 ma, how much current does the device use? Assume an ideal transformer. Real Transformers P s < P P eff = P s / P P a) down b) 6 V c) 3 A Reasons for power losses in real transformers 1. resistance of wires in P and S coils causes heating of coils 2. not all flux from P coil is linked to S coil 3. core warms up as result of cycles of flux changes (hysteresis) Health and Safety Concerns associated with High-Voltage Power Lines 1. Extra-low-frequency electromagnetic fields, such as those produced by electrical appliances and power lines, induce currents within a human body. Just as AC can induce emfs and currents in secondary coils, so to can they be induced in the human body since it is a conducting medium Changing magnetic field induces current in human body 2. Current research suggests that low-frequency fields do not harm genetic material. f = 60 Hz individual photons of this frequency do not have enough energy to cause ionization in the body childhood leukemia clusters are suspected to have a link to living near overhead power cables 3. The risks attached to the inducing of current in the human body are not well-understood. Modern transformers are up to 99% efficient 4. small currents are induced in core (eddy currents) reduce by lamination 17 Risks are likely to be dependent on current density, frequency, and length of exposure 18

Power Transmission Power loss in transmission lines When current flows through a wire, some energy is lost to the surroundings as the wire heats up due to the collisions between the free electrons in the current and the lattice ions of the wire. This is known as Joule heating or resistive heating. Since the energy lost per second, or power loss, is proportional to the square of the current (P = I 2 R), this energy loss is also know as I 2 R loss. Methods of reducing I 2 R loss in power transmission lines 1. Reduce resistance: thicker cables low resistivity material Constraints: lengths are fixed, thicker cables are heavier and more expensive 2. Increase voltage: step voltage up to very high levels Constraints: high voltages are dangerous must be stepped back down for household use For economic reasons, there is no ideal value of voltage for electrical transmission. Typical values are shown below. 1. AC power is generated at a power plant at 12,000 V and then stepped up to 240,000 V by step-up transformers. 2. The high-voltage, low-current power is sent via high-voltage transmission lines long distances. 3. In local neighborhoods, the voltage is stepped-down (and current is stepped-up) to 8000 V at substations. 4. This voltage is stepped-down even further at transformers on utility poles on residential streets. An average of 120 kw of power is delivered to a suburb from a power plant that is 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the transmission voltage is a) 240 V a) 240,000 V I = 500 A P = 100 kw I = 0.50 A P = 0.10 W 19