PHYS 1444 – Section 02 Lecture #15Magnetic Forces on a Moving ChargeExample 27 – 3Path of charged particle in B FieldExample 27 – 4Cyclotron FrequencyTorque on a Current LoopSlide 8Magnetic Dipole MomentMagnetic Dipole Potential EnergyThe Hall EffectExample 27 – 8Slide 13Tuesday March 29, 2011 1PHYS 1444-002 Dr. Andrew BrandtPHYS 1444 – Section 02Lecture #15 Tuesday Mar 29 2011Dr. Andrew BrandtHW7 Ch 27 is due Fri. April 1 @10pmHW8 Ch 28 is due Th April 7 @ 10pmHW9 Ch 29 is due Tu April 12 @ 10 pmReview April 12Test 2 will be Thurs April 14 on Ch 26-29Chapter 27 •Magnetic Force+TorqueTuesday March 29, 2011 2PHYS 1444-002 Dr. Andrew Brandt•This can be an alternative way of defining the magnetic field.–How?–The magnitude of the force on a particle with charge q moving with a velocity v • •What is ?–The angle between the magnetic field and the direction of particle’s movement•When is the force maximum?–When the angle between the field and the velocity vector is perpendicular.• Magnetic Forces on a Moving ChargesinF qvB q=maxF qvB=–The direction of the force follows the right-hand-rule and is perpendicular to the direction of the magnetic fieldmaxFBqv=Tuesday March 29, 2011 3PHYS 1444-002 Dr. Andrew BrandtExample 27 – 3 Magnetic force on a proton. A proton having a speed of 5x106m/s in a magnetic field feels a force of F=8.0x10-14N toward the west when it moves vertically upward. When moving horizontally in a northerly direction, it feels zero force. What is the magnitude and direction of the magnetic field in this region? What is the charge of a proton? What does the fact that the proton does not feel any force in a northerly direction tell you about the magnetic field?Why? pq =The field is along the north-south direction. Because the particle does not feel any magnetic force when it is moving along the direction of the field. Since the particle feels force toward the west, the field should be pointing to …. North Using the formula for the magnitude of the field B, we obtainB =Fqv=1419 68.0 100.101.6 10 5.0 10 /NTC m s--�=״�We can use a magnetic field to measure the momentum of a particle. 191.6 10e C-+ = �Tuesday March 29, 2011 4PHYS 1444-002 Dr. Andrew Brandt•What is the shape of the path of a charged particle moving in a plane perpendicular to a uniform magnetic field?–Circle!! Why?–An electron moving to right at the point P in the figure will be pushed downwardPath of charged particle in B Field–At a later time, the force is still perpendicular to the velocity–Since the force is always perpendicular to the velocity, the magnitude of the velocity is constant–The direction of the force follows the right-hand-rule and is perpendicular to the direction of the magnetic field–Thus, the electron moves in a circular path with a centripetal force F.Tuesday March 29, 2011 5PHYS 1444-002 Dr. Andrew BrandtExample 27 – 4 Electron’s path in a uniform magnetic field. An electron travels at a speed of 2.0x107m/s in a plane perpendicular to a 0.010-T magnetic field. Describe its path. What is the formula for the centripetal force? Since the magnetic field is perpendicular to the motion of the electron, the magnitude of the magnetic force isSince the magnetic force provides the centripetal force, we can establish an equation with the two forces F =( ) ( )( )( )31 72199.1 10 2.0 101.1 101.6 10 0.010kg m smC T---״�= �� �F =F =r =Solving for rma =m2vrevBevB =2vmrmveB=Tuesday March 29, 2011 6PHYS 1444-002 Dr. Andrew Brandt•The time required for a particle of charge q moving w/ constant speed v to make one circular revolution in a uniform magnetic field, , is•Since T is the period of rotation, the frequency of the rotation is•This is the cyclotron frequency, the frequency of a particle with charge q in a cyclotron accelerator–While r depends on v, the frequency is independent of v and r. Cyclotron FrequencyT =B v^rr12qBfT mp= =2 rvp=2vp2 mqBpmvqB=Tuesday March 29, 2011 7PHYS 1444-002 Dr. Andrew Brandt–The magnetic field exerts a force on both vertical sections of wire.–Where is this principle used?•Ammeters, motors, volt-meters, speedometers, etc•The two forces on the different sections of the wire exert a net torque in the same direction about the rotational axis along the symmetry axis of the wire.•What happens when the wire turns 90 degrees?–It will not rotate further unless the direction of the current changes Torque on a Current Loop•What do you think will happen to a closed rectangular loop of wire with electric current as shown in the figure?–It will rotate! Why?Tuesday March 29, 2011 8PHYS 1444-002 Dr. Andrew Brandt Torque on a Current Loop•Fa=IaB•The moment arm of the coil is b/2–So the total torque is the sum of the torques by each of the forces •Where A=ab is the area of the coil–What is the total net torque if the coil consists of N loops of wire?–If the coil makes an angle w/ the field•So what would be the magnitude of this torque?–What is the magnitude of the force on the section of the wire with length a?t =NIABt =sinNIABt q=2bIaB2bIaB+ =IabB =IABTuesday March 29, 2011 9PHYS 1444-002 Dr. Andrew Brandt Magnetic Dipole Moment–It is a vector•Its direction is the same as that of the area vector A and is perpendicular to the plane of the coil consistent with the right-hand rule–Your thumb points to the direction of the magnetic moment when your finger cups around the loop in the direction of the wire–Using the definition of magnetic moment, the torque can be written in vector form•The formula derived in the previous page for a rectangular coil is valid for any shape of the coil•The quantity NIA is called the magnetic dipole magnetic dipole moment of the coilmoment of the coilNIAm=rrNIA B Bt m= � = �rr rr rTuesday March 29, 2011 10PHYS 1444-002 Dr. Andrew Brandt Magnetic Dipole Potential Energy•Where else did you see the same form of torque?–Remember the torque due to electric field on an electric dipole? –The potential energy of the electric dipole is – •How about the potential energy of a magnetic dipole?–The work done by the torque is– –If we chose U=0 at /2, then C=0–Thus the potential energy is•Very similar to the electric dipolecosU B Bm q m=- =- �rrt =rU =U =p E�rrp E- �rrdt q =�sinNIAB dq q =�cosB Cm q- +Tuesday March 29, 2011 11PHYS 1444-002 Dr. Andrew Brandt•This is called the
View Full Document
Unlocking...