Magnetic Field and ForcesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces1Magnetic PolesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces2 If a bar-shaped permanentmagnet, or bar magnet, isfree to rotate, one endpoints north. This end iscalled a north pole or Npole; the other end is asouth pole or S pole. Opposite poles attracteach other, and like polesrepel each other.Magnetism and certain MetalsMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces3 Either pole of a permanent magnet will attract a metallike iron.Magnetic Field of the EarthMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces4The earth itself is a magnet.Magnetic MonopolesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces5Magnetic monopolesdo not exist.Electric Currents and MagnetsMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces6 In 1820, Hans Oersteddiscovered that a current-carrying wire causes acompass to deflect. This discovery revealed theconnection between movingcharges and magnet.The Idea of a Magnetic FieldMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces7Magnetic Interactions arise in 2 stages:1. A moving charge or a current produces a magnetic fieldin the surrounding space (in addition to its electric field).2. This magnetic field exerts a force on any other movingcharges or current present in the field.Magnetic Field Is a vector field and assigned the symbol For any magnet, points out of its north pole and into itssouth poleMagnetic Force Ms. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces8 The magnetic force onq is perpendicular toboth thevelocity of q and the magnetic field. The magnitude of the magnetic force is = sin.Magnetic Force Ms. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces9 The magnetic force onq is perpendicular toboth thevelocity of q and the magnetic field. The magnitude of the magnetic force is = sin.Magnetic Force Ms. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces10 The magnetic force onq is perpendicular toboth thevelocity of q and the magnetic field. The magnitude of the magnetic force is = sin.Magnetic Force Ms. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces11 The magnetic force onq is perpendicular toboth thevelocity of q and the magnetic field. The magnitude of the magnetic force is = sin. The direction of the field is identified by the right handrule. The magnetic force on a moving charged particle, both inmagnitude and direction, is given by = Magnetic Force Ms. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces12 Two charges of equal magnitude but opposite signsmoving in the same direction in the same field willexperience magnetic forces in opposite directions.Measuring Magnetic FieldsMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces13 A cathode-ray tube can be used to determine thedirection of a magnetic field.Check Your UnderstandingMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces14Two particles, having the same charge but differentvelocities, are moving in a constant magnetic field. Whichparticle, if either, experiences the greater magnetic force?Check Your UnderstandingMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces15Four particles follow the paths shown in the figure asthey pass through a magnetic field. What are thecharges of each particle?+--Check Your UnderstandingMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces16An electron moving in the direction of the +z-axis entersa magnetic field. If the electron experiences a magneticdeflection in the +y direction, the direction of themagnetic field in this region points in the direction of thea. x-axisb. + z-axisc. y-axisd. z-axise. + y-axisExample 1. Magnetic Force on a ProtonMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces17A proton moves at a speed of 5.0106m/s along the +x-axis and encounters a magnetic field whose magnitude is0.40 T directed at an angle of =30.0 with respect to theprotons velocity. Find (a) the magnitude and direction ofthe magnetic force on the proton and (b) the acceleration ofthe proton. The mass of a proton is 1.67x10-27kg.Answer:(a) Magnitude: FB= 1.6x10-13NDirected 90 to both v and B following the right-hand rule.(b) a = FB/m = 9.6x1013m/s2.Magnetic Field LinesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces18 The lines originate from the north pole and end on the south pole; theynever start or stop anywhere and always go around in closed loops. The magnetic field at any point is tangent to the magnetic field line atthat point. The strength of the field is proportional to the number of lines per unitarea that passes through a surface oriented perpendicular to the lines. The magnetic field lines never intersect.Magnetic Field LinesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces19Magnetic FluxMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces20 The magnetic flux through a surface is defined in the sameway as the electric flux is defined.The magnetic flux through anarea element is

= = cos

=

The total magnetic flux througha surface

= Magnetic FluxMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces21 Magnetic flux is a scalar quantity. In the special case in which B is uniform over a planesurface with total area A, and are the same at allpoints on the surface, and

= = If the magnetic field happens to be perpendicular to thesurface, then

= The SI unit of magnetic flux is the weber.Magnetic FluxMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces22

= =

=

= = = Gauss Law for MagnetismMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces23 The total magnetic flux through a closed surface is zero. = Example 2. Magnetic FluxMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces24The magnetic field in a certain region of space is 0.385 T, directedalong the +x-axis. In the shaded volume shown, calculate the netmagnetic flux through the enclosing surface by finding the flux througheach surface.xzy40 cm30 cm30 cmSurfaces at top, bottom and surface at x-y plane: FB= 0 since their area vectors are perpendicular to B.Surface at y-z plane: cos180(0.385T)(0.3m)( 0.4)0.046BB A BAmWbF Front rectangular surface:cos(36.9)(0.385T)(0.3m)(0.5m)(0.8)0.046WbBB A BA F Net magnetic flux = 0.Magnetic Forceon a Current-Carrying ConductorMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces25Force on each charge:

=

Total # of charges in the wire: = Total force on wire segment: =

=

= Magnetic Forceon a Current-Carrying ConductorMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces26If the field is not perpendicularto the wire but makes an angle = In general, the magnetic force ona current-carrying conductor isgiven by =

Check Your UnderstandingMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces27A current-carrying wire is placed in the same magneticfield B in four different orientations (see figure). Rank theorientations according to the magnitude of the magneticforce exerted on the wire in decreasing order.1. B and D2. A3. CExample 3. Magnetic Force on a WireMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces28A straight, horizontal wire carries a current of 28A directed out ofthe page. If the wires linear mass density is 46.6g/cm, what arethe magnitude and direction of the magnetic field required tofloat the wire (i.e., to balance its weight)?Answer:To balance the wire, the magnetic force must be equal to the wires weight:So the magnitude of the magnetic field isIlB mg 3 22( / ) (46.6 10 kg/m)(9.8m/s )1.6 10 T28Am l gBI Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces29 The net force on a current loop in a uniform magnetic fieldis zero. But the net torque is not, in general, equal to zero. Torque is maximum when is parallel to the plane of theloop or perpendicular to the vector normal to the plane ofthe loop. =

= = = sin

= =

2 sin90 = Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces30 When is not perpendicular to the vector normal to theplane of the loop but makes an angel with respect to it =

= = cos = 0 = sin = Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces31 Torque is zero when is perpendicular to the plane of theloop or parallel to the vector normal to the plane of theloop. = Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces32 Magnetic dipole moment = where is the area vectornormal to the loop The direction of is definedto be perpendicular to theplane of the loop, with asense determined by theright-hand rule.Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces33 The magnitude of torque ona current loop is given by = The torque can then beexpressed as = = The torque tends to rotatethe loop in the direction ofdecreasing , towards itsstable equilibrium position.Force and Torque on a Current LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces34 Torque on a Solenoid = where is the # of turnsExample 4. Torque on Current-Carrying LoopMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces35A 22cm by 35cm rectangular loop of wire carrying a current of 1.40A isoriented with its plane perpendicular to a uniform 1.50T magnetic field.Calculate the net force and net torque on the loop.Answer:Net force is zero.Net torque is zero.The Direct-Current (DC) MotorMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces36m aligns with B: rotor turns ccwm now aligned with B: rotor srillturns ccw because of inertiam aligns with B: rotor turns ccwExample 5. Magnetic Force on ChargesMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces37Each of the labeled points at the corners of the cube in the figurerepresents a charge of magnitude q=1.0nC moving with a velocity inthe specified direction with magnitude v=3.0x105m/s. A magnetic fielddirected along the +x-axis has magnitude 0.80T. Find the magnitudeand direction of the force on each charge and indicate the direction onthe diagram.++--BxzyExample 6. Current in the RodMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces38A horizontal rod 0.200m long is mounted on a balance andcarries a current. At the location of the rod there is auniform horizontal magnetic field with magnitude 0.087Tdirected perpendicular to the rod. The magnetic force onthe rod is measured by the balance to be 0.22N. How muchcurrent is in the rod?Answer:I = F / LBsinf = (0.22) / (0.2)(0.087)sin90 = 12.6 AExample 7. Magnetic Force on a ChargeMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces39A uniform magnetic field with magnitude 1.2mT is directed verticallyupward throughout the volume of a chamber. A proton with kineticenergy 5.3MeV enters the chamber and moves horizontally fromsouth to north. What is the magnitude and direction of the magneticforce on the proton?Answer:The force is dependent on the speed of a charged particle whichcan be solved from its kinetic energy K = mv2:So the magnetic force isdirected west to east.6 13 27 72 2(5.3 10 eV)(1.602 10 J/eV) 1.67 10 kg 3.2 10 m/s v K m 19 7 2 3 15sin (1.602 10 C)(3.2 10 m/s )(1.2 10 T)sin90 6.1 10 NBF qvB 6 19 27 72 2(5.3 10 eV)(1.602 10 J/eV) 1.67 10 kg 3.2 10 m/s v K m Example 8. Torque on Current-Carrying SolenoidMs. Arlyn D. Macasero Chap 27 - Magnetic Field and Magnetic Forces40A solenoid has a loop area of 2.0104m2, consists of 100 loops orturns, and contains a current of 0.045 A. The coil is placed in a uniformmagnetic field of magnitude 0.15 T. (a) Determine the magneticmoment of the solenoid. (b) Find the maximum torque that themagnetic field can exert on the coil.Answer:(a) Magnetic moment:

= = 100 0.045 A2.0 104 m2 = 0.9 103 A m2(b) Maximum torque (when f = 90):

=

sin90 = 0.9 103 A m20.15 T

= 0.135 103 N m