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 34
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 35
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 36
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 37
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 38
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 PHY2054: Chapter 19 39
Visual of B Field Around Wire PHY2054: Chapter 19 40
B Field Example I = 500 A toward observer. Find B RHR field is counterclockwise ( 7 π ) μ 4 10 500 0i 0.0001 B = = = 2π r 2π r r 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 41
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 42
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 43
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 44
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 ( R /2) 2 A enc π ienc = i = i = A 2 tot π R i 4 R r R B s B i Δ = = 1 μ0i B = B 22π R i 2π μ0 2 4 μ i 0 = On surface 2π R PHY2054: Chapter 19 45
Ampere s Law (cont) Same problems: use Ampere s law to solve for B at any r Wire radius R, total current i i i i i 2 2 Aenc π r r enc = = = A 2 2 tot π R R B Δ s = μ i i 0enc R 2 r B ( 2 ) i Δ s = B πr = μ0i 2 R μ0i r B = r R 2π R R r B = μ i 0 2π r r R PHY2054: Chapter 19 46
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 47
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 48
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 49
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 50
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 51
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 52