Magnetic Forces & Fields

1st & 2nd Hand Rules ConceptsSuppose we placed several compasses around a long,straight wire. If we were to run a substantial amount ofcurrent through the wire, a magnetic field much strongerthan Earth’s would be produced, and the compassneedles would orient themselves to point along a circlecentred on the wire.1.a) A solenoid (long, narrow coil) isconnected to a battery as shown.Copy the diagram and illustratethe direction of electron flow withan arrow. (Electrons are repelledfrom the battery’s negativeterminal and attracted to itspositive terminal).b) Determine which end of thesolenoid is north and which issouth. Label these poles on yourdiagram. Which Hand Rule did you use? The magnetic field generated by a long, straight wire hasthe shape of circles centred on the wire. The directionsof the magnetic field vectors are clockwise when weview the field so that the electrons are flowing toward us( ), and counter-clockwise when viewed with theelectrons flowing away ( ).Side Viewc) Illustrate the dipolar magnetic field produced by thesolenoid.d) What direction would a compass point if it was placed atpoint “P” in the diagram? What direction would a compasspoint if it was placed at point “Q” in the diagram?2.Two solenoids are positioned as illustrated.a) Copy the diagram andlabel the direction ofelectron flow in the coilswith arrows. Determinewhich end of each coil isthe north pole and whichis south. Label the poles.Top Viewb) What is the direction ofthe magnetic fieldproduced at point “P” bythe solenoid on the upper left? Draw and label this field asB1 on the diagram. c) What is the direction of the magnetic field produced atpoint “P” by the solenoid on the lower right? Draw andlabel this field as B2 on the diagram.First Hand Rule: Point your left thumb in the directionof the flow of electrons (or other negative sources).Your fingers will curl in the direction of the magneticfield loops surrounding the source current. For positivesources, use your right hand instead.We often use loops or coils of wire to produce magneticfields. The magnetic field of a coil comes out of one end(called the north pole), loops around the outside of thecoil, and re-enters the coil through its south pole. Thistype of magnetic field is called dipolar d) Assuming that as B1 and B2 are equal in strength, whatis the direction of the net magnetic field at point “P”? I.e.which way would a compass needle at point “P” point?Draw a vector diagram below to illustrate.3.An electron in an atom orbits around the nucleus in thedirection shown in the diagram.Which side of the atom (top orbottom) becomes the northpole of the magnetic fieldproduced by the electron?Explain.4.a) The illustration shows a wire with the direction ofelectron flow indicated. Which hand rule would you use todetermine the magnetic field produced by the wire?b) Illustrate on a copy of the diagram the direction that acompass would point at points “P” and “Q”. Second Hand Rule: Curl your left fingers in thedirection of the electron flow. Your thumb will point tothe north end of the coil. For positive sources, use yourright hand.PHYSICS 30 (Unit 2B)c) What is the direction of B directly in front of the wire?What is the direction of B directly behind the wire?3

Magnetic Force (3rd Hand Rule) Gravitational and electric forces do not depend on the velocitiesof the test mass or charge. The direction of these forces issimply the direction of the field unless the test charge isnegative, in which case the force is directed opposite to thefield.Concepts1.a) For each solenoid, determine where the northand south poles are located (i.e. top or bottom),and whether the magnetic field along the axispoints up or down. Illustrate your answer on acopy of the diagram and label the magnetic field asB. Which hand rule did you use?Magnetic forces are more complicated: the direction of theforce depends on whether the test charge is positive or negative,and it also depends on the direction of the test charge’svelocity.b) For each current-carrying wire in the abovediagram, determine the direction of the magneticforce it experiences. The arrow direction indicatesthe electron flow. On your diagram, label themagnetic force on the wire as Fm. Which hand ruledid you use?Directions of B and Fm withElectron Flow (v) Out of the Page Third Hand Rule: With your thumb pointed in the direction ofthe test charge’s velocity (v) and your fingers extended in thedirection of the magnetic field (B), your palm will be facing thedirection of the magnetic force (Fm). Use your left hand fornegative test charges and your right hand for positive ones. According to this rule, the direction of the magnetic force actingon a test charge will always be perpendicular to both thevelocity of the test charge and the magnetic field (i.e. thedirection that a compass needle would point.) Test charges at rest experience no magnetic force. Moving testcharges experience a magnetic force that increases with thespeed of the test charge. The magnitude of the magnetic force isdirectly proportional to speed:2.Contrast this equation to the simpler ones for gravity (Fg mg)and electric (Fe qE) forces. The symbol v means the component of the velocity that isperpendicular to the magnetic field: v v sin θ, where θ is theangle between v and B. The parallel component of the velocity is not relevant to theforce because the 3rd Hand Rule says there is no force when thevelocity is parallel to B! c) Determine the direction of the magnetic field thatthe left wire experiences due to the current in theright wire. Label this direction B. Determine thedirection of the magnetic force acting on the leftwire. Label this Fm.3.where I is the current flowing through the wire and L is thelength of wire experiencing (the perpendicular component of)the magnetic field B.PHYSICS 30 (Unit 2B)An electron passesthrough a parallel platecapacitor as shown.a) What is the direction ofthe electric force (Fe) thatthe capacitor exerts onthe electron? Draw and label this direction on acopy of the diagram.b) A scientist wants to create a magnetic force (Fm)on the electron in the opposite direction as theelectric force. What direction should the magneticfield (B) have? Draw the directions of B and Fm onyour diagram.We often need to know the magnetic force not just on a singletest charge, but also on all of the charges moving through awire. The force for this situation is Fm ILB a) Two long, straight, parallel wires carry currentsas shown. Determine the direction of the magneticfield that the right wire experiences (i.e. this is thetest current) due to the current in the left wire(source). Label this direction B on a copy of thediagram. Which hand rule did you use?b) Determine the direction of the magnetic forceacting on the right wire. Label this Fm. Which handrule did you use? Fm qv B In the two scenarios illustrated below, consider thesolenoids as the source of the magnetic field, andthe wires as the test current experiencing themagnetic force.4.An electron is traveling toward the top of this page,through a magnetic field directed into the page.Draw and label the directions of v and B below.Determine the direction of Fm and label it. Sketchthe resulting motion of the electron.4

CalculationsProblems5.A wire carrying 1.25 A of current experiences a force of 3.25mN when held perpendicular to the magnetic field betweenthe poles of a horseshoe magnet. If the length of wire withinthe magnetic field is 5.20 cm, what is the average magneticfield strength? Include a diagram with the directions of v, Band Fm labelled.7.An alpha particle (see formula sheet) moving with a4speed of 9.20 10 m/s passes through a uniform2.05 mT magnetic field and experiences a force of–173.50 10 N. What angle does the particle’s velocitymake with the magnetic field?8.6.A wire with 720 mA of current (electron flow east) passesthrough a magnetic field of 50.0 mT (down). A total lengthof 35.0 cm of wire is exposed to the magnetic field.Determine the total magnetic force (including direction) thatthe wire experiences. Include a diagram with the directionsof v, B and Fm labelled. A wire experiences an upward magnetic force of48.2 mN when held so that its current flows north.When the wire is turned so that its current flows east,the magnetic force is 60.5 mN downward. In whatdirection does the uniform, horizontal magnetic fieldpoint?Lab Activity: Magnetic Forces & FieldsExperiment 1: Magnetic ForceExperiment 2: Magnetic Field of a SolenoidIn this experiment, you will use an apparatus similar to theone illustrated below to test the hypothesis that the magnetic force on a section of wire is Fm ILB .Design and conduct your own experiment to study themagnetic field produced by a solenoid electromagnet.Explicitly state which variable you will manipulate, whichones you will control and which will be responding. Youdesign must include a hypothesis and a detailed procedure.Your analysis must allow you to reach a conclusion as towhether or not your hypothesis is supported by the data.A copper wire is mounted to a lightweight frame and placedon a balance, which is then zeroed. A horseshoe magnet ismounted to a lab stand so that one of its poles its directlybeside (but not touching) the wire, producing a horizontalmagnetic field. The current running through the wire ismanipulated using a rheostat and measured with anammeter. The force that the magnet exerts on the wire is theresponding variable: it is measured by converting the “mass”reading from the balance to a “weight” in Newtons. Also,measure the length of wire within the magnetic field.Graph the data and determine a model equation. Doesthe model equation support the hypothesis? Use theequation to estimate the magnetic field strength adjacent tothe horseshoe magnet’s pole.PHYSICS 30 (Unit 2B)5

Motion in a Magnetic Field We saw in the electricity unit that moving test charges ina uniform electric field will experience a constant force,resulting in projectile motion: motion along a parabolictrajectory. Fnet ma When a test charge moves in a magnetic field, the forcehas a magnitude of Fm qv B with a direction given bythe 3rd Hand Rule.If the magnetic field is uniform (and there are no otherforces besides the magnetic force), the magnitude of theforce will remain constant but, according to the 3rd HandRule, the direction of the force must change so that it isalways perpendicular to v. mv qBAurora BorealisIf the initial velocity is not perpendicular to B, theperpendicular component of the velocity (v ) will executeuniform circular motion while the parallel component (v )will be unaffected by the field. The resulting trajectorywill have the shape of a helix centred on a magnetic fieldline (i.e. the field line forms the axis of the helix).Helical Motion(Negative Test Charge)We can solve the resulting equation to obtain theradius of the uniform circular or helical motion r We can substitute the appropriate equations for themagnetic force (Fnet qv B) and centripetalacceleration (a v 2/r) v 2qv B mrThe resulting motion will be uniform circular motion ifthe initial velocity is perpendicular to the magnetic field!Uniform Circular Motion(Positive Test Charge with v B)Newton’s 2nd Law determines the motion of the testcharge The aurora borealis (Northern Lights) and auroraaustralis (Southern Lights) are lights that appear in thesky at polar latitudes. These lights are produced whencharged particles from the Sun (known as the solarwind) collide with molecules in Earth’s upperatmosphere.NASA image from sec.gsfc.nasa.gov/popscise.jpg. The reason these lights occur near the poles has to dowith Earth’s and the Sun’s magnetic fields. Magneticfield lines from Earth’s poles form large loops thatconnect up with the Sun’s magnetic field. Solar windparticles move in helixes around these field lines fromthe Sun to Earth’s poles. The magnetic field loops near Earth’s equator aremuch smaller: they loop back to Earth without everconnecting to the Sun’s field. Since they are nomagnetic field lines connecting the Sun to Earth’sequatorial regions, solar wind particles rarely reachEarth’s atmosphere in the tropics.This image was produced using the King’s Centre for Visualization inScience simulation located at:http://www.kcvs.ca/site/projects/physics files/particleMField3d/particleME3d.swfPHYSICS 30 (Unit 2B)6

Cyclotron A particle accelerator is a device used to acceleratesubatomic particles (e.g. protons) to extremely highspeeds. One type of particle accelerator is a linear accelerator(linac). This consists of a series of electrodes alternatingbetween positive and negative. In 1929, Ernest Lawrence invented the cyclotron.Like a linac, a cyclotron used the electric fieldbetween two oppositely charged plates to acceleratethe particles. The cyclotron, however, also uses a magnetic field tomove the particles along a circular trajectory, so thatthe same pair of electrodes can accelerate the particlesrepeatedly.CyclotronLinear Accelerator Electrodes A proton (for example) would accelerate away from thefirst ( ) electrode toward the (–) one. At the instant whenthe proton passes through the negative plate, the polarityis reversed so that the proton will keep accelerating. Thisprocess repeats with the polarity reversing every time theproton passes through an electrode. As the particles gain speed, the circumference of theirtrajectory increases proportionally. This means thatthe effective distance between electrodes increasesautomatically, so that the timing for reversing thepolarity of the plates remains constant!A problem with linacs is that as the proton gains speed,the distance between the electrodes must be increased. Toreach extreme speeds, linacs must be several hundredmetres long!88-Inch Cyclotron at the University of California at html)PHYSICS 30 (Unit 2B)7

Calculations1.Simulation: Motion in a Magnetic Field6An electron moving with a speed of 5.20 10 m/s enters a regionwhere the magnetic field strength is 12.9 mT. The velocity isperpendicular to the magnetic field.a) Determine the magnetic force that the electron experiences.b) Determine the centripetal acceleration and the radius of theelectron’s orbit if its velocity is perpendicular to the magnetic field.2.A proton from the solar wind moves through Earth’s6magnetosphere with a speed of 3.00 10 m/s. If the magnetic field–5strength is 4.00 10 T and makes a 20.0 angle with the proton’svelocity, what will the radius of the proton’s helical motion be?Problems3.Calculate the period of circular motion for a proton moving in acyclotron with a magnetic field strength of 0.750 T. Determine thespeed of the proton when it leaves the cyclotron (r 30.0 cm).4.A cyclotron has a radius of 42.0 cm and uses a magnetic fieldstrength of 1.20 T.Pre-LabIn this activity, you will use the King’s Centre forVisualization in Science (KCVS) simulation tostudy the motion of an electron moving in uniformcircular motion through a uniform magnetic field.Before beginning the simulation, derive therelationship between the magnetic field strengthand the radius of the resulting circular motion.Which variables would need to be controlled in areal experiment? Is the relationship linear? If not,what transformation would you need to perform tolinearize a graph of your data?Simulation1.Open the link to the KCVS particle motionsimulation from the Magnetism unit on yourteacher’s webpage.2.Under the magnetic field control, select the zcomponent (Bz) to simulate a magnetic fielddirected into the page. Click the “#” button forthe magnetic field and set it to 0.005 Teslas.Make a note of the electron velocity. Click“Play” and observe the uniform circularmotion of the electron.3.When the electron has completed an entireorbit, click “Pause”. Next click “Measure”.Click a point on the circle’s circumference onthe left side. Then click a point on the circlefrom the right side. Finally, click a point onthe circle midway between the two pointsalready marked. The simulation will measurethe radius of the circle. Record the magneticfield strength and the radius in a table. Clickyour mouse once more to erase themeasurement.c) What is the mass of the ion? What two elements could the ionbe? [Hint: How many times heavier is the ion than a proton?]4.Click “Reset”, and then increase the magneticfield Bz by 0.005 T. Click “Play”.d) What is the ion’s speed?5.a) A proton executes uniform circular motion due to a magneticfield. If the magnetic field strength was doubled, by what factorwould the proton’s orbital radius change?Repeat steps 3 and 4 until you have about 10different radii for 10 different magnetic fieldstrengths.6.Transform your data to make it linear. Graphthe linearized data and add a best-fit line.7.Determine the equation of the best-fit line.8.Does the data support the relationshipbetween B and r that you derived in the PreLab? By substituting your experimental (bestfit) equation into the theoretical (derived)equation, calculate the mass of the electron.(Use the electron charge of q 1e –191.60 10 C.) Does the calculated massagree with the accepted value of the electronmass?a) Calculate the cyclotron period for protons, i.e. the period ofprotons in circular motion within the cyclotron.b) Calculate the speed and the kinetic energy of a proton about toleave the cyclotron.c) If the potential difference between the plates was 2.50 kV, howmany revolutions did the proton need to make before leaving thecyclotron? [Hint: Remember that the potential difference tells youthe work done on the protons each time they pass through theaccelerating plates.]5.A positive ion with a net charge of one elementary charge unit isaccelerated from rest through a 11.0 kV potential difference. Theion then moves into a region where the magnetic field strength is1.05 T and follows a circular trajectory with a radius of 25.0 mm.The magnetic field is perpendicular to the velocity.22a) Explain why qV ½mv and why qvB mv /r?b) Eliminate v from these two equations and solve for m.6.b) A proton executes uniform circular motion due to a magneticfield. If an alpha particle with the same speed as the proton wereplaced in the same magnetic field, by what factor would its orbitalradius differ from the proton’s orbit?7.Derive an equation for the “cyclotron period”, the time for onecircular orbit of a particle with charge q and mass m moving in aperpendicular, uniform magnetic field B. Show that this period doesnot depend on the speed of the particle or the radius of itstrajectory.–14–8Answers: (1) 1.07 10 N, 2.30 mm, (2) 268 m, (3) 8.74 10 s, (4)–873–2755.47 10 s, 4.83 10 m/s, 2.43 10 , (5) 5.01

the field corresponds to the density of field lines. A unique property of magnetic fields is that there are no points of convergence or divergence: magnetic field lines always form closed loops with no beginning or end! Concepts 1. Examine the following field diagrams to determine which is a magnetic fie