Chapter 29: Creating Magnetic Fields PHY2049: Chapter 29 1 Creating Magnetic Fields ÎSources of magnetic fields Spin of elementary particles (mostly electrons) Atomic orbits (L > 0 only) Moving charges (electric current) ÎCurrents generate the most intense magnetic fields Discovered ÎThree by Oersted in 1819 (deflection of compass needle) examples studied here Long wire Wire loop Solenoid PHY2049: Chapter 29 2 B Field Around Very Long Wire ÎField around wire is circular, intensity falls with distance Direction given by RHR (compass follows field lines) µ 0i B= 2π r µ0 = 4π ×10−7 Right Hand Rule #2 PHY2049: Chapter 29 3 Long Wire B Field Example ÎI = 500 A toward observer. Find B vs r RHR ⇒ field is counterclockwise µ i ( 4π ×10 ) 500 0.0001 B= = = −7 0 2π r r r r r r r r = = = = = = = 0.001 m 0.005 m 0.01 m 0.05 m 0.10 m 0.50 m 1.0 m 2π r B B B B B B B = = = = = = = r 0.10 T 0.02 T 0.010 T 0.002 T 0.001 T 0.0002 T 0.0001 T = = = = = = = 1000 G 200 G 100 G 20 G 10 G 2G 1G PHY2049: Chapter 29 4 Charged Particle Moving Near Wire ÎWire carries current of 400 A upwards moving at v = 5 × 106 m/s downwards, 4 mm from wire Find magnitude and direction of force on proton Proton ÎSolution of force is to left, away from wire Magnitude of force at r = 4 mm Direction µ0 I F = evB = ev 2 π r ( F = 1.6 × 10−19 )( −7 2 10 × × 400 6 5 × 10 0.004 ) F = 1.6 × 10−14 N PHY2049: Chapter 29 v I 5 Ampere’s Law ÎTake arbitrary path around set of currents Let ienc be total enclosed current (+ up, − down) Let B be magnetic field, and ds be differential length ∫ B ⋅ ds = µ0ienc ÎOnly along path Not included in ienc currents inside path contribute! 5 currents inside path (included) 1 outside path (not included) PHY2049: Chapter 29 6 Ampere’s Law For Straight Wire ÎLet’s try this for long wire. Find B at distance at point P Use circular path passing through P (radius r) From symmetry, B field must be circular ∫ B ⋅ ds = B ( 2π r ) = µ0i B= ÎAn µ 0i 2π r P r easy derivation PHY2049: Chapter 29 7 Useful Application of Ampere’s Law ÎFind B vs r inside long wire, assuming uniform current Wire radius R, total current i Find B at radius r = R/2 ÎKey ienc fact: enclosed current ∝ area R π ( R / 2 )2 i = = i 2 πR 4 r R i B 2π = µ0 2 4 1 µ 0i B= 2 2π R µ0i B= 2π R PHY2049: Chapter 29 On surface 8 Force Between Two Parallel Currents ÎForce on I2 from I1 µ0 I1I 2 µ0 I1 F2 = I 2 B1L = I 2 L= L 2π r 2π r RHR ⇒ Force towards I1 ÎForce on I1 from I2 µ0 I1I 2 µ0 I 2 F1 = I1B2 L = I1 L= L 2π r 2π r RHR ⇒ Force towards I2 ÎMagnetic I2 I2 I1 forces attract two parallel currents I1 PHY2049: Chapter 29 9 Force Between Two Anti-Parallel Currents ÎForce on I2 from I1 µ0 I1I 2 µ0 I1 F2 = I 2 B1L = I 2 L= L 2π r 2π r RHR ⇒ Force away from I1 ÎForce on I1 from I2 µ0 I1I 2 µ0 I 2 F1 = I1B2 L = I1 L= L 2π r 2π r RHR ⇒ Force away from I2 ÎMagnetic I2 I2 I1 forces repel two antiparallel currents I1 PHY2049: Chapter 29 10 Parallel Currents (cont.) ÎLook B at them edge on to see B fields more clearly B 2 Antiparallel: repel 1 F 2 1 F B 2 1 B Parallel: attract F 1 2 F PHY2049: Chapter 29 11