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Experiment No. 6 Solenoid and Electro-Magnetism

Experiment No. 6 Solenoid and Electro-Magnetism Reference: Physics by Wilson&Buffa, (Ch. 19.2, 19.4, 19.6, 19.7) 2016/2017. Solenoid and Electro-magnet Applications Magnetic Field Strength and Magnetic Force (Ch 19.2)

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Experiment No. 6 Solenoid and Electro-Magnetism

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  1. Experiment No. 6 Solenoid and Electro-Magnetism Reference: Physics by Wilson&Buffa, (Ch. 19.2, 19.4, 19.6, 19.7) 2016/2017 • Solenoid and Electro-magnet Applications • Magnetic Field Strength and Magnetic Force (Ch 19.2) • Define magnetic field strength and the magnetic force exerted by a magnetic field on a moving charge. • Magnetic Forces and Current Carrying Wire (Ch. 19.4) Calculate the magnetic force on the current carrying wire. • What has the Oersted experiment taught us? • Electromagnetism: The Source of Magnetic Fields (Ch. 19.6) • Solenoid Experiment Test No. 1 • Calculation of magnetic field in simple cases • Magnetic Materials and Magnetic Permeability (CH. 19.7) • Electromagnet and Magnetic Permeability or how to enhance the magnetic field strength • Solenoid experiment Test No. 2 • Solenoid Experiment Lessons Learned • New Terminology • Quiz No. 6 • Homework No. 6

  2. Solenoid and Electro-magnet + - A wire-wound solenoid forms a tube 2 1/2 inches long, and ¼ inch in diameter. An iron nail is inserted part way into the solenoid. Then, when an electric current is applied to the solenoid by closing the switch, a magnetic field is created inside the coil and the nail moves! 2

  3. Solenoid Applications Transducer devices that convert energy into motion Solenoid Valve- an electromechanical solenoid which actuates either a pneumatic or hydraulic valve or a solenoid switch Sprinkler valve solenoid Automobile starter solenoid- This type of solenoid switch uses lighter current-carrying wire and it is used in cars to turn on the electric starter motor switch which uses much higher current and heavy wires.

  4. Devices that use Electromagnets Electromagnets are used in many situations where an easily and quickly variable magnetic field is desired, for example: Electromagnets provide beam deflection for cathode ray tubes & mass spectrometers. Electromagnets are essential in many circuit-breakers. They are used in cars in electromagnet brakes and clutches. In some trains, electromagnetic brakes grip directly onto the rails. Electromagnets are used in cranes to lift heavy scraps of iron and steel. Electromagnets are used in shipyard cranes to lift heavy containers. Electromagnets separate metals at junkyards and recycling centers. Experimental magnetic levitation trains use powerful electromagnets to hover without touching the track. Electromagnets are used in rotary electric motors to produce a rotating magnetic field that turns the rotor. (Millions are produced.) 4

  5. Industrial lifting magnet

  6. Magnetic Field Strength and Magnetic Force (Ch. 19.2) Define magnetic field strength and determine the magnetic force exerted by a magnetic field on a moving charge particle. • Experiments indicate there is a connection between electrical properties of objects and how they respond to magnetic fields. Electromagnetism is a study of interactions between electrically charged particles and magnetic fields. • Consider the horseshoe magnet in the presence of a moving electric charge q+.There must exist a centripetal force that is perpendicular to the particle’s velocity to cause its particle’s velocity deflection. An electric force is not present. The gravitational force is too weak. Therefore, the force is due to the interaction of the moving charge and the magnetic field, B. B = F/qv [T (Tesla) = N/Am] F = qvB (v perpendicular to B) F = qvB sin θ θ Note: 1 T, N/Am is too large for most applications. 1 G (Gauss) = .0001T; Earth’s magnetic field = .0 to3G; Superconducting material = 25T

  7. Magnetic Forces and Current Carrying Wire (Ch. 19.4) Calculate the magnetic force on the current carrying wire. F = Σq x vB; v = L/t [m/s] F = Σq/t x LB; Σq/t = I [A] F = I x LB; [ N= AxmxN/Am] F = ILB sin θ; Electric current is composed of many electric charges. Each electric charge moving in a magnetic field is acted on by a force unless it is moving parallel to or directly opposite the direction of the field. θ

  8. Exercise 19.4: Magnetic forces on wire suspended at the equator Near the Equator it is theoretically possible to cancel the pull of gravity with an upward magnetic force on a wire. What must be a) the current direction and b) the current magnitude? a) Use of the Right–Hand rule: Point F up and B north. The resulting current, I direction (or its velocity v) is East. b) Collect the data using the SI units to calculate the current using the equation F=ILB; Given: B, Earth’s magnetic field: .4G; B = .4Gx(10-4T/G) = .4x10-4T Wire length, L: 1m Wire mass, m: 30gr; m= 30gr = 3x10-2 kg Wire weight or force, w = mg w = 3x10-2 kg x 9.8m/s2 = 29.5x10-2 kgm/s2 = 29.5x10-2 N Current, I = F/LB = 29.5 N/(1mx.4x10-4T) = 72.5x10+4 A I = 7.25 x10+3 A (Not practical!)

  9. Oersted discovery how current in a long straight wire is creating a magnetic field I N S N S W Compass positionsd2d1 Magnetic Field The curled hand indicates the circular sense of the magnetic field direction. It is tangent to the circular field lines at any point. The compass is oriented North to South in line with the current flow in a wire. When undisturbed by the local magnetic fields the compass aligns its self with Earth’s magnetic field which has a strength of 0.00005T or 5 x 10-5T. (Ref: Physics 19.2) The electric current’s induced magnetic field strength must exceed that of the Earth to make the compass to move. μ0 x I B, magnetic flux density = ----------- ; (Ref.: Physics eq. 19-12) Test No. 1, d1 = 1 cm 2 π d Test No. 3, d3 = 10 cm Assuming that the D-cell can deliver 10 -15 A max B1 = 20x10-5T B2 = 2x10-5T In the test no. 1 the magnetic strength B was high enough to make the compass move while in the test no. 3 with the wire 10 to 15 cm away from the compass it wasn’t strong enough to overcome Earth’s magnetic field strength. (Magnetic permeability of free space μ0 = 4 π x 10-7 N/A2 )

  10. About today’s experiments • When current (I) flows through a wire a small magnetic field (B) is created. • When a current flows through a coil of wire, all of the individual wires’ magnetic fields join together to make a stronger magnetic field. In today’s experiment we shall try to demonstrate what physicist Ampere discovered long ago how by having two or more such wires the electric current through all should interact magnetically and strengthen the magnetic field which should exert even stronger force that causes movement. • We’ll first calculate the magnetic field strength generated by an electric current flowing through a coil of wire.

  11. Magnetic Field in a Current Carrying Solenoid L Current moving out of the page I = .33 A Cross-section through the center of the core It is constructed by winding a long wire into a tight coil with many circular loops in a free space where the magnetic permeability, μ0 is equal to 4 π x 10-7 Tm/A . V = 6 VDC If the solenoid has N turns (loops) and carries a current, I where the length of a wire is L the magnetic field strength, B is: μ0 N I B = -------------- where L = 2.5 in (coil exposed) = .063m; N = 50; V = 6 V; I = .33A L (Ref: Physics eq. 19.14) B = 55 x 10-5 T (11 times stronger than the Earth’s magnetic field 5 x 10-5) The magnetic field created by a solenoid is proportional to both the number of turns in the winding, N, and the current in the wire, I, hence the product, NI, is called ampere-turns.

  12. Solenoid Demonstration + - A wire-wound solenoid forms a tube 2 1/2 inches long, and ¼ inch in diameter. An iron nail is inserted part way into the solenoid. Then, when an electric current is applied to the solenoid by closing the switch, a magnetic field is created inside the coil and the nail moves! Is there a way to increase solenoid’s magnetic field strength so that it could lift an object? 12

  13. Magnetic Permeability The symbol μ, represents magnetic permeability, the measure of the ability of a material to support the formation of magnetic field of its core material. It is used in electromagnets and other electrical equipment (motors, generators, transformers and consumer electronics). A core of a large ferromagnetic material with a large permeability in an electromagnet can enhance its field thousands of times as compared with the free space. μ N I B = -------------- ; L = length of the coil 0.063 m; N = number of turns 50; L I = current 10A; μ = permeability for electrical steel is μ = 5x10-3 Tm/A (Ref. Physics eq. 19.14) Use of ferromagnetic material magnifies the magnetic field permeability.

  14. Electromagnet and Magnetic Permeability I = 10A It is constructed by winding a long wire into a tight coil with many circular loops in and around a bar of steel with the magnetic permeability, μequal to k x μ0. (see Wikipedia on permeability) V = 6 VDC If the solenoid has N turns (loops) and carries a current, I where the length of the coil is L than the magnetic field strength, B is: μ x N x I B = -------------- ; L = 2.5 in = .063m; N = 50; I = 10A; μ = 5x10-3 Tm/A L (Ref: Physics eq. 19.14) B = 397 x 10-5 T (80 times stronger than the Earth’s magnetic field 5 x 10-5) The magnetic field created by a solenoid can be enhanced by the magnetic permeability of the core material with higher permeability than the free space.

  15. Exercise, Ch. 19.4: Magnetic Forces on Current Carrying Wire A 2 m long current carrying wire of 20A in a magnetic field of 50mT whose direction is at an angle of 370 from the direction of the current. Find the force on the wire. F I I 530 B Ix θ Vertical current component to the magnetic field axis, Ix is applied to determine the projected current vector value perpendicular to the Magnetic B. Ix = I x sin530 = 20 A x .62; L= 2m; B = 50mT = 20x10-3T; F = IxLBsinθ F = 20 x.62 x 2 x 50x10-3 [AmTx(N/AmT)]; Reminder: B=F/qv [T=N/Am] F = 1.24 N

  16. Exercise Ch 19.4 Force of an Electromagnet L I = .33 A What happens to the length of a spring when a large amount of current passes through it? Magnetic force must overcome the force of the spring (F=kx) and the spring shortens or lengtens? Magnetic force, F = iLB ; B = (μ0 x I)/(2 π d)(Ref.: Physics eq. 19-12)It shortens!

  17. How to make the Solenoid? Materials needed: 5 ft length of #22 magnet wire 2.5 inch plastic straw Two jumper wires (1” bare wires) Push-Button switch Small piece of sandpaper Small finishing nail To Build the Coil: Place the magnet wire at one end of the plastic tube, leaving about 1” over the straw for connecting the coil. Wind 10 turns of the wire tightly over the tube, then press the turns together to place them close. Repeat this winding and compressing 5 times (50 turns total). Be sure to leave about 1” free at the end for connection. (see picture next slide) Using small piece of sandpaper, strip the insulation (red colored coating) from about ½” of each end of the coil. The bare wire should be copper colored.

  18. Prepare the Breadboard to Conduct Solenoid Experiment Place the push-button switch with one terminal at location e2, and the other terminal at g4. (see picture) Press down on the switch until it’s flat on the board. Bend one of the bare jumpers into a “U” shape and insert it between a2 and the RED buss. Bend the other bare jumper into a “U” and insert it between j22 and the BLUE buss. Your breadboard is now ready for the coil (see picture) Completing the Breadboard. Insert the leads of the coil, one into location j4, and the other into i22. Connect the Positive (+) 6V terminal of the battery pack to the RED buss jumper using the RED alligator clip lead. Connect the Negative (-) terminal of the battery pack to the BLUE buss jumper using the BLACK alligator clip lead. Insert the nail into the switch end of the tube so aprox. ½” sticks out. You are now read for the test.

  19. Jumper 1: a2 to RED Bus Push-button Switch: e2 to g4 To j4 Breadboard Connections Ends sanded to expose copper Coil – 50 turns Jumper 2: j22 to BLUE Bus To i22 Plastic Tube

  20. To Battery (+) 6 VDC Connections to the Battery Jumper 2 (-) 0 VDC Jumper 1 (+) 6 VDC To Battery (-) 0 VDC

  21. Conduct the Test No. 1 • Make sure everything is connected according to the test setup instructions. • Test No. 1: Insert the test nail into the sleeve half way. Turn ON the the Push-button Switch. The nail should quickly slide into the sleeve! • You have just witnessed a solenoid switch in action! • Turn OFF the toggle switch!

  22. Conduct the Test No. 2 • Insert the 2 ½ in long steel bar inside the wire coil that is supported by the straw tube. Tighten the wire ends around it. • Test No. 2: Press the push button switch. The nail should be attracted by the solenoid which is acting as an electromagnet! • You have just witnessed an electromagnet in action! • Turn OFF the push button switch! (The current level is very high when the electric circuit is closed.)

  23. Solenoid Test Data Sheet February 2017 Name: _________________________ Test No. 1: When the push-button is pressed, the nail will be attracted into the coil by the magnetic forces produced by the current flowing in the coil. Question No. 1: Why has the nail moved after the solenoid was energized?_______________________________________________________________________________________ Q. No. 2: Why is the nail pulled into the coil, rather than be ejected from it?_______________________________________________________________________________________ Q. No. 3: Does an electro-magnet work on Non-ferrous materials? (example Gold, or Aluminum)? _______________________________________________________________________________________ Q. No. 4: Could the solenoid magnetic field strength be increased by coiling the wire around a ferromagnetic material? _______________________________________________________________________________________ Test No. 2: Q. No. 5: Why was the solenoid able to attract a nail or a paper clip after inserting the steel bar into the core of the solenoid?_________________ _________________________________________________________________________ Q. No. 6: Did the core steel bar remain magnetized after turning OFF the electric current to the solenoid?___________; Was the steel bar made of a “hard” or “soft” ferromagnetic material?________ Explain the difference between the two materials’ characteristics: _______________________________________________________________________________________

  24. Solenoid Experiment Lessons Learned • An electric current flowing through a straight piece of wire 30 inch long can generate sufficient magnetic field strength to overcome that of the Earth as demonstrated by the moving of a compass. • We have also learned that by coiling the wire tightly next to each other we can magnify significantly the magnetic field strength proportionally to the number of turns (loops), N in the winding and the current in the wire, I. The product NI is called Ampere turns, At. This term is commonly used to express the strength of a solenoid. • Solenoid’s magnetic field can become much stronger when winding the coil around a bar of a high permeability material such as electric steel or permalloy. Material’s permeability is defined by μ = kμ0 where k is the material permeability relative factor.

  25. New Terminology • Magnet- It produces an invisible magnetic field that creates a force that pulls on other ferromagnetic materials such as iron and that attracts or repels other magnets. • Ferromagnetic- A ferromagnetic material is one that has magnetic properties similar to those of iron. Once magnetized it retains magnetic properties. • Permanent magnet- It retains a permanent orbital motion of electrons (“spins” similar to Earth). All magnetism is due to circulating electric currents. In magnetic material the magnetism is produced by electrons orbiting within the atoms; in most substances the magnetic effects of different electrons cancel each other out, but in some, such as iron, a net magnetic field can be induced by aligning the crystalline structure of atoms. • Magnetic field- It is created by an electric current (electromagnet) or the by an already permanently magnetized material spinning motion of electrons (permanent). A magnetic field strength, H (Amps per meter) is a mathematical description of the magnetic influence of electric currents and magnetic materials. • Magnetic Permeability- the measure of the ability of a material to support the formation of a magnetic field. It is a degree of magnetization that a material obtains in response an applied magnetic field. B = μH, magnetic induction flux density, B measured in Tesla or Gauss and specified by direction and magnitude. μ0 is permeability of the free space; μ is permeability of a ferromagnetic material; B= F/qv ; F is the force of motion, N; q is the electric charge, C; v is velocity, m/s; • Ampere- Ampere deduced from Oersted’s experiment that electric current in a wound coil exhibits all properties of an “electromagnet”. • Electromagnet- Magnetic field that is created by the movement of electrons in an electrical field. • Solenoid- The term refers specifically to a long, thin loop of wire, often wrapped around a metallic core, which produces a uniform magnetic field in a volume of space. • Ampere-turns- A magneto-motive force of a current of one Ampere flowing in a single-turn loop.

  26. SI Electromagnetism Units Symbol Name Unit SI Conversion I Electric current Ampere A = C/s q Electric charge Coulomb C = As E Electromotive force, Potential difference Volt J/C = (kgms2/As ρResistivity Resistivity m P Electric power Watt W = kgm/s2 E Electric field strength Volt per meter V/m H Magnetic field strength Ampere per meter A/m B Magnetic flux density, Induction (F/qv) Tesla T = Wb/ms2 μ Permeability henry per meter H/m = kgm/(As) μ0 Permeability of free space 4 π x 10-7N/A2 F Force of motion N 1 N = 1 kg m/s2 v Velocity m/s Resistance Ohm V/A = 1 Ω = 1 kg·m2·s-3·A-2 in Inch Inch 0.0254 m or 25.4 mm T Tesla Tesla Vs/m2 = N/Am = Wb/ms2 N Newton Newton Kgm/s2 Wb Weber Weber 1 V·s = 1 T·m2 = 1 J/A; J Joulekgm2/s2 ; 2.78 x 10-7 kWh; 2.39 x 10-4 kcal; 9.48 x10-4 BTU

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