Magnetic Forces and Torque on Current Loops - Prof. Benjamin T. Hall, Quizzes of Physics

Download Magnetic Forces and Torque on Current Loops - Prof. Benjamin T. Hall and more Quizzes Physics in PDF only on Docsity! PHY2054: Chapter 19 25 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 I B I is parallel to B, so no magnetic force PHY2054: Chapter 19 26 a a bb Torque on Current Loop Consider rectangular 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 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 ( )Fd iBa b iBab iBAτ = = = = Plane normal is ⊥ B here PHY2054: Chapter 19 29 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 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 FFv R v B qF v PHY2054: Chapter 19 30 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 Circular Motion of Positive Particle B qF v 2mv qvB R = mvR qB = PHY2054: Chapter 19 31 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. 21 2 2KK mv v m = ⇒ = 2mv mKR eB eB = = ( )( )6 19 13 27 10 1.6 10 =1.6 10 J 1.67 10 kg K m − − − = × × = × ( ) 1 2 2 / 2 v v eBf T R mv eB mπ π π = = = = 760Hzf = 2900mR = Frequency is independent of v! PHY2054: Chapter 19 34 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 Particle moves in a circle (if ) ( ) 0K F x F v tΔ = ⋅Δ = ⋅ Δ = v B⊥ PHY2054: Chapter 19 35 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 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 mvR qB = Bigger mass means bigger radius Mass Spectrometer Photographic plate ml Velocity selector ‘© 2006 Brooks/Cole - Thomson PHY2054: Chapter 19 39 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 40 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 x z y Forces cancel on opposite sides of loop PHY2054: Chapter 19 41 Hall Effect: Do + or – Charges Carry Current? + charges moving counter- clockwise 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): This type of experiment led to the discovery (E. Hall, 1879) that current in conductors is carried by negative charges up driftF qv B= down induced H VF qE q w = = drift "Hall voltage"HV v Bw= = PHY2054: Chapter 19 44 B Field Around Very Long Wire Field around wire is circular, intensity falls with distance Direction given by RHR (compass follows field lines) 0 2 iB r μ π = 7 0 4 10μ π −= × Right Hand Rule #2 μ0 = “Permeability of free space” PHY2054: Chapter 19 45 Visual of B Field Around Wire PHY2054: Chapter 19 46 B Field Example I = 500 A toward observer. Find B vs r RHR ⇒ field is counterclockwise 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 ( )7 40 4 10 500 10 2 2 iB r r r πμ π π − −× = = = PHY2054: Chapter 19 49 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 An easy derivation ( ) 0 0 2 2 i B s B r i iB r π μ μ π Δ = = = ∑ r P PHY2054: Chapter 19 50 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 2 enc enc 2 tot 4 A r ii i i A R π π ⎛ ⎞ = × = × =⎜ ⎟⎜ ⎟ ⎝ ⎠ 0 0 2 2 4 1 2 2 i R iB s B iB R π μ μ π ⎛ ⎞Δ = =⎜ ⎟ ⎝ ⎠ = ∑ r R 0 2 iB R μ π = On surface 0 enc i B s iμΔ =∑ r = R/2 PHY2054: Chapter 19 51 Ampere’s Law (cont) Same problems: use Ampere’s law to solve for B at any r Wire radius R, total current i 2 2 enc enc 2 2 tot A r ri i i i A R R i π π ⎛ ⎞ = × = × =⎜ ⎟⎜ ⎟ ⎝ ⎠ = ( ) 2 0 02 0 2 or 2 i rB s B r i i R i rB R R π μ μ μ π ⎛ ⎞ Δ = = ⎜ ⎟⎜ ⎟ ⎝ ⎠ = ∑ r R 0 2 iB r μ π = 0 enc i B s iμΔ =∑ (r ≤ R) r ≥ R (r ≥ R) (r ≤ R) PHY2054: Chapter 19 54 Parallel Currents (cont.) Look at them edge on to see B fields more clearly Antiparallel: repel F F Parallel: attract F F B BB B 2 1 2 2 2 1 1 1 PHY2054: Chapter 19 55 B Field @ Center of Circular Current Loop Radius R and current i: find B field at center of loop Direction: RHR #3 (see picture) If N turns close together 0 2 iB R μ = 0 2 N iB R μ = From calculus PHY2054: Chapter 19 56 Current Loop Example i = 500 A, r = 5 cm, N=20 ( )( )70 20 4 10 500 1.26T 2 2 0.05 iB N r πμ −× = = = ×