Magnetic Force A vertical wire carries a current and is in a vertical magnetic field. What is the direction of the force on the wire? (a) left (b) right (c) zero (d) into the page (e) out of the page B I is parallel to B, so no magnetic force I PHY2054: Chapter 19 25
Consider rectangular current loop Torque on Current Loop Forces in left, right branches = 0 Forces in top/bottom branches cancel No net force! (true for any shape) But there is a net torque! Bottom side up, top side down (RHR) Rotates around horizontal axis τ = Fd = ( iba) b = ibab = iba μ = NiA magnetic moment (N turns) True for any shape!! Direction of μ given by RHR Fingers curl around loop and thumb points in direction of μ b a B b a Plane normal is B here PHY2054: Chapter 19 26
General Treatment of Magnetic Moment, Torque μ = NiA is magnetic moment (with N turns) Direction of μ given by RHR θ Torque depends on angle θ between μ and B τ = μbsinθ PHY2054: Chapter 19 27
Torque Example A 3-turn circular loop of radius 3 cm carries 5A current in a B field of 2.5 T. Loop is tilted 30 to B field. B 30 ( ) 2 2 2 μ = NiA = 3iπr = 3 5 3.14 0.03 = 0.0339A m τ = μbsin30 = 0.0339 2.5 0.5 = 0.042 N m Rotation always in direction to align μ with B field PHY2054: Chapter 19 28
Trajectory in a Constant Magnetic Field A charge q enters B field with velocity v perpendicular to B. What path will q follow? Force is always velocity and B Path will be a circle. F is the centripetal force needed to keep the charge in its circular orbit. Let s calculate radius R x x x x x x x x x x x x x x B x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x vx x v F F R F v q PHY2054: Chapter 19 29
Circular Motion of Positive Particle x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 2 mv mv qvb R = R = qb B F PHY2054: Chapter 19 30 v q
Cosmic Ray Example Protons with energy 1 MeV move earth B field of 0.5 Gauss or B = 5 10-5 T. Find radius & frequency of orbit. 1 2 2 K = mv v = 2K m K m ( )( ) 6 19 13 = 10 1.6 10 =1.6 10 J 27 = 1.67 10 kg mv 2mK R = = eb eb R = 2900m f = 1 v v eb T = 2π R = 2 π mv/ eb = f = 2πm 760Hz ( ) Frequency is independent of v! PHY2054: Chapter 19 31
Helical Motion in B Field Velocity of particle has 2 components v = v + v (parallel to B and perp. to B) Only v = v sinφ contributes to circular motion v = v cosφ is unchanged So the particle moves in a helical path v is the constant velocity along the B field v is the velocity around the circle v v φ v B R = mv qb PHY2054: Chapter 19 32
Helical Motion in Earth s B Field PHY2054: Chapter 19 33
Magnetic Field and Work Magnetic force is always perpendicular to velocity Therefore B field does no work! Why? Because Consequences Kinetic energy does not change Speed does not change Only direction changes Δ K = F Δ x = F vδ t = v B Particle moves in a circle (if ) ( ) 0 PHY2054: Chapter 19 34
Magnetic Force Two particles of the same charge enter a magnetic field with the same speed. Which one has the bigger mass? A B Both masses are equal Cannot tell without more info x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x R = mv qb Bigger mass means bigger radius x x x x x x x x x x x x x x x x x x x x x x x x A B PHY2054: Chapter 19 35
Mass Spectrometer PHY2054: Chapter 19 36
Mass Spectrometer Operation Positive ions first enter a velocity selector where E B and values are adjusted to allow only undeflected particles to enter mass spectrometer. Balance forces in selector select v qe = qvb v= E/ B 2 Spectrometer: Determine mass from v and measured radius r 1 r r 1 2 mv = 1 qb mv = 2 qb PHY2054: Chapter 19 37
Mass Spectrometer Example A beam of deuterons travels right at v = 5 x 10 5 m/s What value of B would make deuterons go undeflected through a region where E = 100,000 V/m pointing up vertically? ee = evb 5 5 B = E/ v = 10 /5 10 = 0.2T If the electric field is suddenly turned off, what is the radius and frequency of the circular orbit of the deuterons? mv R f 2 ( 27 )( 5 3.34 10 5 10 ) mv 2 = evb R = = = 5.2 10 m eb 1 v 5 10 6 = = = = 1.5 10 Hz T 2π R 6.28 5.2 10 5 ( )( 2 ) ( 19 1.6 10 )( 0.2) PHY2054: Chapter 19 38
Quiz: Work and Energy A charged particle enters a uniform magnetic field. What happens to the kinetic energy of the particle? (1) it increases (2) it decreases (3) it stays the same (4) it changes with the direction of the velocity (5) it depends on the direction of the magnetic field Magnetic field does no work, so K is constant PHY2054: Chapter 19 39
Magnetic Force A rectangular current loop is in a uniform magnetic field. What direction is the net force on the loop? (a) +x (b) +y (c) zero (d) x (e) y B Forces cancel on opposite sides of loop z x y PHY2054: Chapter 19 40
Hall Effect: Do + or Charges Carry Current? + charges moving counterclockwise experience upward force Upper plate at higher potential charges moving clockwise experience upward force Upper plate at lower potential Equilibrium between magnetic (up) & electrostatic forces (down): V Fup = qvdriftb down = H induced = drift F qe q w V = v Bw= "Hall voltage" H This type of experiment led to the discovery (E. Hall, 1879) that current in conductors is carried by negative charges PHY2054: Chapter 19 41
Electromagnetic Flowmeter E Moving ions in the blood are deflected by magnetic force Positive ions deflected down, negative ions deflected up This separation of charge creates an electric field E pointing up E field creates potential difference V = Ed between the electrodes The velocity of blood flow is measured by v = E/B PHY2054: Chapter 19 42
Creating Magnetic Fields Sources of magnetic fields Spin of elementary particles (mostly electrons) Atomic orbits (L > 0 only) Moving charges (electric current) Currents generate the most intense magnetic fields Discovered by Oersted in 1819 (deflection of compass needle) Three examples studied here Long wire Wire loop Solenoid PHY2054: Chapter 19 43
B Field Around Very Long Wire Field around wire is circular, intensity falls with distance Direction given by RHR (compass follows field lines) B = μ i 0 2π r μ0 = 4π 10 7 Right Hand Rule #2 μ 0 = Permeability of free space PHY2054: Chapter 19 44
Visual of B Field Around Wire PHY2054: Chapter 19 45
B Field Example I = 500 A toward observer. Find B vs r RHR field is counterclockwise ( 7 π ) μ 4 10 500 0i 10 B = = = 2π r 2π r r 4 r = 0.001 m B = 0.10 T = 1000 G r = 0.005 m B = 0.02 T = 200 G r = 0.01 m B = 0.010 T = 100 G r = 0.05 m B = 0.002 T = 20 G r = 0.10 m B = 0.001 T = 10 G r = 0.50 m B = 0.0002 T = 2 G r = 1.0 m B = 0.0001 T = 1 G PHY2054: Chapter 19 46
Charged Particle Moving Near Wire Wire carries current of 400 A upwards Proton moving at v = 5 10 6 m/s downwards, 4 mm from wire Find magnitude and direction of force on proton Solution Direction of force is to left, away from wire Magnitude of force at r = 0.004 m μ0i F = evb = ev 2π r F F 7 19 6 2 10 400 ( 1.6 10 )( 5 10 ) = 0.004 14 = 1.6 10 N v I PHY2054: Chapter 19 47
Ampere s Law Take arbitrary path around set of currents Let i enc be total enclosed current (+ up, down) Let B ll be component of B along path B Δ s = μ i i 0enc Only currents inside path contribute! 5 currents inside path (included) 1 outside path (not included) Not included in i enc PHY2054: Chapter 19 48
Ampere s Law For Straight Wire Let s try this for long wire. Find B at distance at point P Use circular path passing through P (center at wire, radius r) From symmetry, B field must be circular i B = An easy derivation 0 2π r ( 2π ) μ0 B Δ s = B r = i μ i r P PHY2054: Chapter 19 49
Useful Application of Ampere s Law Find B field inside long wire, assuming uniform current Wire radius R, total current i Find B at radius r = R/2 Key fact: enclosed current area B Δ s = μ i i 0enc 2 Aenc π r i enc = = = A 2 tot π R 4 i i i r = R/2 R r R B s B i Δ = = 1 μ0i B = B 22πR i 2π μ0 2 4 μ i 0 = On surface 2π R PHY2054: Chapter 19 50
Ampere s Law (cont) Same problems: use Ampere s law to solve for B at any r Wire radius R, total current i B Δ s = μ 0enc i i i i i = i 2 2 Aenc π r r enc = = = A 2 2 tot π R R i (r R) (r R) R r 2 r B ( 2 ) 0 or i Δ s = B π r = μ i μ 2 0i R μ0i r B = 2π R R (r R) B = μ i 0 2π r r R PHY2054: Chapter 19 51
Force Between Two Parallel Currents Force on I 2 from I 1 μ0i1 μ0i1i2 F2 = I2B1L= I2 L L 2πr = 2πr RHR Force towards I 1 Force on I 1 from I 2 μ0i2 μ0i1i2 F1 = I1B2L= I1 L L 2πr = 2πr RHR Force towards I 2 I 2 I 1 Magnetic forces attrac two parallel currents I 2 I 1 PHY2054: Chapter 19 52
Force Between Two Anti-Parallel Currents Force on I 2 from I 1 μ0i1 μ0i1i2 F2 = I2B1L= I2 L L 2πr = 2πr RHR Force away from I 1 Force on I 1 from I 2 μ0i2 μ0i1i2 F1 = I1B2L= I1 L L 2πr = 2πr RHR Force away from I 2 I 2 I 1 Magnetic forces repel two antiparallel currents I 2 I 1 PHY2054: Chapter 19 53
Parallel Currents (cont.) Look at them edge on to see B fields more clearly B 2 1 Antiparallel: repel 2 1 B F F B 2 1 Parallel: attract 2 1 B F F PHY2054: Chapter 19 54
B Field @ Center of Circular Current Loop Radius R and current i: find B field at center of loop μ0i B = From calculus 2R Direction: RHR #3 (see picture) If N turns close together Nμ0i B = 2R PHY2054: Chapter 19 55
i = 500 A, r = 5 cm, N=20 Current Loop Example ( )( 7 ) 20 4 10 500 = = = 1.26T 2 2 0.05 μ0i π B N r PHY2054: Chapter 19 56
Formula found from Ampere s law i = current n = turns / meter B = μ in 0 B ~ constant inside solenoid B ~ zero outside solenoid Most accurate when B Field of Solenoid L R Example: i = 100A, n = 10 turns/cm n = 1000 turns / m B ( 7)( )( 3 π ) = 4 10 100 10 = 0.13T PHY2054: Chapter 19 57