Electromagnetism

4.3 Lab Assignment: Magnetism Name: Matthew Breeding Part 1 Electromagnetic Induction We all use batteries for flashlights, cell phones, and many other devices. We know they deliver energy through electricity. Where do they get their energy? If you are not sure, watch the first video listed in the assignment: Discover How Car Batteries Work . Briefly explain where a battery gets its energy. When electrons start to move rapidly this is called electricity. The car battery can be recharged while the motor is running because of the alternator. If you are not sure what an alternator is or how it works, watch the second video listed in the assignment: How Automotive Alternators Work, made simple An alternator drives current through the battery to recharge it. What kind of device is it and where does it get the energy that is carried by the current into the battery? A rotor spins inside of a coil of wire making an A.C current. The A.C current is then converted to D.C current and sent to the battery While the alternator is complicated, there is a much simpler version of the same device, sometimes called a pick-up coil. For a simulation of a pick-up coil, activate the simulation after clicking on the third link listed in the assignment: Faraday's Law PhET Simulation . Move the magnet around and notice current flow in the coils. While the alternator is complicated, there is a much simpler version of the same device, sometimes called a pick-up coil. For a simulation of a pick-up coil, activate the simulation after clicking on the third link listed in the assignment: Faraday's Law PhET Simulation (Links to an external site.). (If the Simulation is not available, use the video showing a real magnet and pick-up coil to answer the questions.) 1) Position the north end of the magnet closest to the right side of the coil. With the magnet at rest, describe the voltmeter reading. There is no reading because the magnet is not moving. 2) Push the north end of the magnet into the coil. Describe the voltmeter reading while the magnet is in motion. It reads fully negative then fully positive because of the saides of the amgnt pasing through the coil. 3) Leave the north end of the magnet at rest inside the coil. Describe the voltmeter reading. There is no reading once again because the magnet is not moving 4) Remove the north end of the magnet from the coil by pulling it out to the right. 1

Describe the voltmeter reading while the magnet is in motion. The voltmeter reads fully positive because only the north side is being pulled through the coil The challenging thing about such a simulation understanding where the energy is coming from to drive current through the circuit. The following videos will make this clearer. Watch the next video listed in the assignment: Eddy Current Tubes . What force is slowing the magnet’s fall? (Put an X by your choice.) a gravity b electrical X c magnetic d nuclear If you are not sure, watch the next listed video: World's First Electric Generator . Since energy cannot be created or destroyed, where does the energy come from to drive the current in the pipe that generates the magnetic fields that slow the magnet’s fall? By the magnet moving inside the tube it is generating electricity Since energy cannot be created or destroyed, where does the energy go after the magnet has fallen through the tube? It is released Part 2 Forces on Charged Particle in an Electric and Magnetic Fields Electric and magnetic fields exert forces on charged particles. These forces are at the heart of all important electromagnetic inventions, including motors, generators, transformers, and a device used to measure the mass of individual atoms, the mass spectrometer. A: Electric Field Forces 2

Trial 3 1000 N/C 5 x 10 5 m/s +10 μC 32 ns Trial 4 1000 N/C 10 x 10 5 m/s +5 μC 45 ns Note that the greater the force, the less time it should take for the particle to hit the plate, because the same mass has a larger force acting on it, creating a greater acceleration. Based on these results which conclusions can we draw about the effect of the strength of the electric field, the amount of charge, and the velocity of motion on the force exerted on the particle? If the strength of the electric field is increased, a the force on the charged particle decreases. b the force on the charged particle does not change. X c the force on the charged particle increases. If the amount of charge on the particle is increased, a the force on the charged particle decreases. b the force on the charged particle does not change. X c the force on the charged particle increases. If the velocity of the particle is increased, a the force on the charged particle decreases. X b the force on the charged particle does not change. c the force on the charged particle increases. The formula for calculating the force on a charge particle in an electric field is Force = (Charge)(Electric Field Strength) or F = qE B: Magnetic Field Forces Open the following simulation: https://ophysics.com/em7.html 4

At left is the notation used to represent vectors that are oriented so they would come out of the plane of the page or into the plane of the page, perpendicular to the page. Start with the default values of the simulation and record the information listed below: What type of charge is used in the default settings? a positive X b negative In which direction is the magnetic field pointing? a Left, in the plane of the simulation field. b Right, in the plane of the simulation field. c Up, in the plane of the simulation field. X d Down, in the plane of the simulation field. e Coming out of the plane of the simulation (indicated by dots) f Going into the plane of the simulation (indicated by x’s) 5

To make the charge negative, move the charge slider to the left until it indicates a negative number in the field. To make the magnetic field go into the page, move the magnetic field strength slider to the left until it indicates a negative number in the field. You may want to clear the simulation field of the previous trace by hitting the “Clear Trace” button. Type of charge (+ or –) Field Direction (in or out) Initial Force direction (up or down) + out ¤ DOWN + in Ä DOWN – out ¤ UP – in Ä UP Next, we want to experiment to determine what quantities affect the magnitude of the force exerted on the charge by the magnetic field. First, let’s see if the strength of the magnetic field affects the amount of force. The greater the centripetal force, the smaller the radius the circular motion will be. Return your simulation to the default values by reloading the simulation. Change the charge to − 3 × 10 − 16 C . Change the magnetic field strength to 6.3 T Run the simulation and record the radius of the circle (check the box “Show Radius”). Quantity Value and unit Charge -3 x 10 -16 C Magnetic field strength 6.3 T velocity 7.5 x 10^6 m/s radius 2.38 mm Now double the charge and run the simulation again. Result. When the charge is doubled the radius a is doubled X b is cut in half c does not change This means that when the charge increases, the force a decreases X b increases c does not change Hit “Reset” and “Clear Trace,” adjust the simulation to the following values, run the simulation, and record the radius. Quantity Value and unit 7

Charge − 4 × 10 − 16 C Magnetic field strength 5 T velocity 7.5 × 10 6 m s radius 2.25 mm Double the Magnetic field strength and run the simulation. Result. When the field strength is doubled the radius a is doubled X b is cut in half c does not change This means that when the field strength increases, the force a decreases X b increases c does not change There is a third influence on the force on a moving particle in a magnetic field, the velocity, but that is harder to understand using this simulation. The formula for calculating the force ( F ) on a moving particle in a magnetic field is F = qvB q = the charge in Coulombs v = the velocity in m/s B = the magnetic field strength in a unit called a Tesla, which it the field strength which will cause a 1 kg particle with 1 C of charge moving at 1 m/s to experience a force of 1 N. The formula for centripetal force is F = m v 2 r If we set these two force formulas equal and solve for the radius ( r ) we get r = mv qB So, if mass, charge, and magnetic field stay constant and we double the velocity, we expect this to double the radius. Reset, Clear Trace, set the simulation to the following values, run the simulation, and record the radius. Quantity Value and unit Charge − 4 × 10 − 16 C Magnetic field strength 8 T velocity 5.0 × 10 6 m s radius 0.94 8

X b the force on the atom is perpendicular to the velocity. c the force on the atom is in the same direction as the velocity. The atoms collide with a detector which marks where each atom strikes. The further from the opening the atom gets, the more mass it has. The mass helps researchers identify the type of atom or isotope. Combining all the measurements the mass spectrometer can make, a calculation can be done that calculates the mass of each individual ion that passes through the field. The mass of each particle can be calculated by using the two force equations used in the previous part. F = qvB and F = m v 2 r If we set these two expressions equal, we can solve for the mass of the particle: m v 2 r = qvB So, m = qv B 1 × r v 2 . Simplify by canceling one of the velocity variables ( v ). m = qBr v We will first do a calculation with a known mass. Then students will get data to calculate for an unknown element. Because the mass of an individual atom is a very small fraction of a kilogram, we will convert each calculation to a unit called an atomic mass unit , which is approximately the mass of a single proton or neutron. Scientists have used the mass of the most common isotope of carbon as the standard to set the value of this unit. So we will do a calculation for a simulated measurement of the mass of a carbon atom. Use the values below in the equation m = qBr v to calculate the mass of a carbon atom in kilograms. charge ( q ) 1.602 × 10 -19 C magnetic field ( B ) 0.00500 T radius ( r ) 0.1245 m velocity ( v ) 5000 m/s Mass in kg 2.67 * 10^-30 kg Mass in amu 161 u Convert this to atomic mass units , where 1 amu = 1.6605 × 10 -27 kg. 10

Divide your calculated mass in kilograms by this number. Your answer should be 12.0 (rounded to 3 significant figures). For the last exercise, see the list of values by your name below. These represent data from a test to identify an element from a pair of isotopes measured with a mass spectrometer. Isotopes are atoms with the same number of protons in the nucleus but different numbers of neutrons. The mass spectrometer will get a different reading for each isotope because the extra neutron or neutrons add mass to the nucleus. Since isotopes behave the same in chemical reactions, the mass spectrometer is needed to separate them and identify them. Each student is trying to identify a different element. In each case, the element has just two isotopes. This immediately eliminates other elements which have more or fewer isotopes than 2. By calculating the masses, you can identify the element. This is a common use of the mass spectrometer, to identify elements and isotopes. 1. Use the values from the table for magnetic field (B), velocity (v), and the first radius (r1), plus the value for the charge (q) listed in the table above to calculate the mass of the first isotope in kilograms, using the equation: m = qBr v 2. Use the second radius (r2) and all the same values for the other variables to calculate the mass of the second isotope in kilograms. 3. Convert the masses from kilograms to atomic mass units by dividing each mass by the number of kilograms in 1 amu. 1 amu = 1.6605 × 10 -27 kg. 4. Consult the table below that lists the established values of the mass of elements with just two isotopes and match your masses with one of them to identify which element your data represents. 5. Do an Internet search to identify one interesting fact about this element. Student B (T) v (m/s) r1 (m) r2 (m) Amin, Zahedullah 0.0050 8000 0.2322 0.2488 Anderson, Shane 0.0070 3500 0.2589 0.2640 Breeding, Matthew 0.0030 8000 0.2792 0.3043 Brooks, Brandon 0.0085 2000 0.2949 0.2997 Cotto, Sebastien 0.0070 3000 0.2795 0.2884 Ellison, Julius 0.0095 1500 0.2863 0.2879 Ferry, David 0.0050 4000 0.2900 0.3065 Genovese, Jacob 0.0080 2500 0.2750 0.2815 Leon, Bailey 0.0095 1100 0.2833 0.2845 Melvin, Blake 0.0075 3000 0.2858 0.2941 Myers, Morgan 0.0085 2200 0.2868 0.2922 Puttkammer, Travis 0.0080 2000 0.2926 0.2977 Short, Nick 0.0015 9000 0.1876 0.2490 Simmons, Tommy 0.0090 1800 0.2859 0.2880 Thompson, Davis 0.0045 1500 0.2727 0.2796 11

See table below for masses of isotopes. Element Isotope Relative Atomic Weight (amu) Element Isotope Relative Atomic Weight (amu) Helium He-3 3.016 Silver Ag-107 106.905 He-4 4.003 Ag-109 108.905 Lithium Li-6 6.015 Indium In-113 112.904 Li-7 7.016 In-115 114.904 Boron B-10 10.100 Antimony Sb-121 120.904 B-11 11.009 Sb-123 122.904 Nitrogen N-14 14.003 Lanthanum La-138 137.907 N-15 15.000 La-139 138.906 Chlorine Cl-35 34.969 Lutetium Lu-175 174.941 Cl-37 36.966 Lu-176 175.943 Vanadium V-50 49.947 Tantalum Ta-180 179.947 V-51 50.941 Ta-181 180.948 Copper Cu-63 62.930 Thallium Tl-203 202.972 Cu-65 64.928 Tl-205 204.974 Gallium Ga-69 68.926 Polonium Po-209 208.982 Ga-71 70.925 Po-210 209.923 Bromine Br-79 78.918 Astatine At-210 209.987 Br-81 80.916 At-211 210.987 Rubidium Rb-85 84.912 Neptunium Np-236 236.047 Rb-87 86.909 Np-237 237.048 13