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Ch. 29 •Certain objects and circuits produce magnetic fields •Magnetic fields, like electric fields, are vector fields •They have a magnitude and a direction •Denoted by B, or B(r) •They have no effect on charges at rest •They produce a force on moving charges given by FB qv B •Perpendicular to magnetic field F q vB sin •Perpendicular to velocity •Magnetic field strengths are measured in units called a tesla, abbreviated T •A tesla is a large amount of magnetic field B F qv N C m/s N T Am Warmup 12 F qv B The Right Hand Rule To figure out the direction of magnetic force, use the following steps: 1. Point your fingers straight out in direction of first vector v 2. Twist your hand so when you curl your fingers, they point in the direction of B vB 3. Your thumb now points in the direction of v B 4. If q is negative, change the sign B •Vectors in the plane are easy to draw v •Vectors perpendicular to the plane are hard •Coming out of the plane •Going into the plane ˆi v B det vx B x ˆj vy By kˆ vz v y Bz vz By ˆi vz Bx vx Bz ˆj Bz vx By v y Bx kˆ Finding the direction proton What is the direction of the force for each F qv B of the situations sketched? A) B) C) D) E) F) None of the above B p v F qv B F q vB sin Ca+2 4. If q is negative, change the sign ion electron e v Ca B v B JIT Quick Quiz 29.1 An electron moves in the plane of this paper toward the top of the page. A magnetic field is also in the plane of the page and directed toward the right. The direction of the magnetic force on the electron is described by which of the following? (a) toward the top of the page (b) toward the left edge of the page (c) toward the bottom of the page (d) toward the right edge of the page (e) upward out of the page (f) downward into the page Ans e Ex- (Serway 29-9) A proton moves perpendicular to a uniform magnetic field B at 1.00 x 10 7m/s and experiences an acceleration of 2.00 x 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field. Solve on Board Warmup 12 Work and magnetic fields How much work is being done on a point charge moving in a magnetic field? F qv B W F s •Work = force distance •Divide the distance into little tiny steps, divide by time dW ds F F v F v cos 0 dt dt •But recall F v • = 90 and cos = 0 Magnetic fields do no work on pure charges F B q v Warmup 13 Cyclotron Motion Consider a particle of mass m and charge q moving in a uniform F qv B magnetic field of strength B B v v F F F q F •Motion is uniform circular motion •Centripetal force formula: mv 2 mv qRB qvB F R p qRB v v •Let’s find how long it takes to go around: 2 R T v 2 m T qB 2 T qB m Motion in a Magnetic Field •The particle may also move parallel to the magnetic fields •No force F qv B •Combined motion is a helix q •Net motion is along magnetic field lines B CT –1 Cosmic rays (atomic nuclei stripped bare of their electrons) which come from all directions, would continuously bombard Earth’s surface if most of them were not deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a magnetic dipole, the intensity of cosmic rays bombarding its surface is greatest at the A. poles. B. mid-latitudes. C. equator. The Earth has magnetic field lines Charged particles from space follow them Hit only at magnetic poles aurora borealis aurora australis Concept Question Two particles with the same mass are moving in the same magnetic field, but particle X is circling in less time than particle Y. What can account for this? A) Particle X is moving faster (only) B) Particle Y is moving faster (only) C) Particle X has more charge (only) D) Particle Y has more charge (only) E) A and C could both account for this F) B and C could both account for this •Moving faster doesn’t help •Higher speed means bigger radius •Higher charge does help •You turn corners faster mv qRB 2 m T qB B X Y Velocity Selector / Mass Spectrometer •When we have both electric and magnetic fields, the force is •Magnetic field produces a force on the charge F q E v B •Add an electric field to counteract the magnetic force •Forces cancel if you have the right velocity 0 E vB v E B FB v FE detector + – Now let it move into region with magnetic fields only •Particle bends due to cyclotron motion •Measure final position •Allows you to determine m/q mv qRB m RB 2 q E m RB q v An electron has a velocity of 1.00 km/s (in the positive x direction) and an acceleration of 2.001012 m/s2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 15.0 N/C (in the positive z direction), determine the components of the magnetic field. If a component cannot be determined, enter 'undetermined'. F q E v B E vB F ma E vB q q 31 12 2 9.1094 10 kg 2.00 10 m/s kˆ 1.602 1019 C v B 26.37kˆ 15.0kˆ v B 11.37kˆ By .02637 T ˆ ˆ ˆ i j k 3 3 ˆ ˆ 10 kBy ˆjBz 26.37k det 10 0 0 Bz 0 B B B y z B undetermined x x The Hall effect •Consider a current carrying wire in a magnetic field •Let’s assume it’s actually electrons this time, because it usually is I FB d V vd t •Electrons are moving at an average velocity of vd I AJ td nqvd •To the left for electrons •Because of magnetic field, electrons feel a force upwards vd I •Electrons accumulate on top surface, positive charge on bottom tdnq •Eventually, electric field develops that counters magnetic force 0 F q EH v d B EH v d B •This can be experimentally measured as a voltage VH EH d vd Bd IBd tdnq IB VH tnq Force on a Current-Carrying Wire •Suppose current I is flowing through a wire of cross sectional area A and length L •Think of length as a vector L in the direction of current •Think of current as charge carriers with charge q and drift velocity vd B I F Nqv d B Vnqv d B ALJ B IL B L F IL B F •What if magnetic field is non-uniform, or wire isn’t straight? •Divide it into little segments dF I ds B •Add them up B I A B F I ds B A JIT Quick Quiz 29.3 A wire carries current in the plane of this paper toward the top of the page. The wire experiences a magnetic force toward the right edge of the page. The direction of the magnetic field causing this force is which of the following? (a) (b) (c) (d) in the plane of the page and toward the left edge in the plane of the page and toward the bottom edge upward out of the page downward into the page Ans c Ex- (Serway 29-35) A wire having a mass per unit length of 0.500 g/cm carries a 2.00 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? Solve on Board Warmup 13 Need to have assumed uniform Force/Torque on a Loop •Suppose we have a current carrying loop in a constant magnetic field •To make it simple, rectangular loop size L W F IL B L •Left and right side have no force at all, because cross-product vanishes I Ft •Top and bottom have forces W ˆ t Ft ILBk Fb B Fb ILBkˆ •Total force is zero •This generalizes to general geometry F I ds B I ds B 0 •There is, however, a torque on this loop τ r F Wˆj Ft ILWBˆj kˆ IABˆi t IAB •What if the loop were oriented differently? •Torque is proportional to separation of forces t FW sin BIWL sin BIA sin Wsin Torque on a Loop (2) F n̂ t W B F Edge-on view of Loop ˆ B τ IAn This formula generalizes to other shapes besides rectangles? •It is true for circular loops, or oddly shaped loops of current Torque and Energy for a Loop ˆ B •Define A to be a vector perpendicular to the loop τ IAn with area A and in the direction of n-hat ˆ A An •Determined by right-hand rule by current A •Curl fingers in direction current is flowing R •Thumb points in direction of A τ IA B I •Define magnetic dipole moment of the loop as τ μB μ IA t B sin •Torque is like an angular force •It does work, and therefore there is energy associated with it U t d B sin d B cos U μ B t A •Loop likes to make A parallel to B B Edge-on view of Loop JIT Quick Quiz 29.4 (i) Rank the magnitudes of the torques acting on the rectangular loops shown below, from highest to lowest. All loops are identical and carry the same current. (ii) Rank the magnitudes of the net forces acting on the rectangular loops shown below, from highest to lowest. (i) c>b>a (ii) a=b=c CT – 2 A rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A. B. C. D. a net force. a net torque. a net force and a net torque. neither a net force nor a net torque. Ans D CT - 3 A rectangular loop is placed in a uniform magnetic field with the plane of the loop parallel to the direction of the field. If a cur-rent is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A. B. C. D. a net force. a net torque. a net force and a net torque. neither a net force nor a net torque. Ans B How to make an electric motor •Have a background source of magnetic fields, like permanent magnets •Add a loop of wire, supported so it can spin on one axis •Add “commutators” that connect the rotating loop to outside wires •Add a battery, connected to the commutators •Current flows in the loop τ IA B •There is a torque on the current loop •Loop flips up to align with B-field •Current reverses when it gets there t •To improve it, make F A the loop repeat many times F τ INA B + –