PHY2049_06-10-11 - 1 Phy2049 Magnetism • Last lecture...

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Unformatted text preview: 6/10/2011 1 Phy2049: Magnetism • Last lecture: Biot-Savart’s and Ampere’s law: – Magnetic Field due to a straight wire – Current loops (whole or bits)and solenoids • Today: reminder and Faraday’s law. hitt Two long straight wires pierce the plane of the paper at vertices of an equilateral triangle as shown. They each carry 3A but in the opposite direction. The wire on the left has the current coming out of the paper while the wire on the right carries the current going into the paper. The magnetic field at the third vertex (P) has the magnitude and direction (North is up): (1) 20 μ T, east (2) 17 μ T, west (3) 15 μ T, north (4) 26 μ T, south (5) none of these X 4 cm Magnetic Field Units • From the expression for force on a current-carrying wire: – B = F max / I L – Units: Newtons/A ⋅ m ≡ Tesla (SI unit) – Another unit: 1 gauss = 10-4 Tesla • Some sample magnetic field strengths: – Earth: B = 0.5 gauss = 0.5 x 10-4 T = 5 x 10-5 T – Galaxy: B ∼ 10-6 gauss = 10-10 T – Bar magnet: B ∼ 100 – 200 gauss – Strong electromagnet: B = 2 T – Superconducting magnet: B = 20 – 40 T – Pulse magnet: B ∼ 100 T – Neutron star: B ∼ 10 8 – 10 9 T – Magnetar: B ∼ 10 11 T Force Between Two Parallel Currents • Force on I 2 from I 1 – RHR ⇒ Force towards I 1 • Force on I 1 from I 2 – RHR ⇒ Force towards I 2 • Magnetic forces cause attraction between two parallel currents I 1 I 2 0 1 0 1 2 2 2 1 2 2 2 I I I F I B L I L L r r μ μ π π = = = I 1 I 2 0 2 0 1 2 1 1 2 1 2 2 I I I F I B L I L L r r μ μ π π = = = Force Between Two Anti-Parallel Currents • Force on I 2 from I 1 – RHR ⇒ Force away from I 1 • Force on I 1 from I 2 – RHR ⇒ Force away from I 2 • Magnetic forces repel two antiparallel currents I 1 I 2 I 1 I 2 0 1 0 1 2 2 2 1 2 2 2 I I I F I B L I L L r r μ μ π π = = = 0 2 0 1 2 1 1 2 1 2 2 I I I F I B L I L L r r μ μ π π = = = Parallel Currents (cont.) • Look at them edge on to see B fields more clearly Antiparallel: repel F F Parallel: attract F F B B B B 2 1 2 2 2 1 1 1 6/10/2011 2 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 2 i B R μ = 2 N i B R μ = Current Loop Example • i = 500 A, r = 5 cm, N=20 ( 29 ( 29 7 20 4 10 500 1.26T 2 2 0.05 i B N r π μ- × = = = × Challenge: How much heat is produced? B Field of Solenoid • Formula found from Ampere’s law – i = current – n = turns / meter – B ~ constant inside solenoid – B ~ zero outside solenoid – Most accurate when L>>R • Example: i = 100A, n = 10 turns/cm – n = 1000 turns / m B in μ = ( 29 ( 29 ( 29 7 3 4 10 100 10 0.13 T B π- = × = Field at Center of Partial Loop • Suppose loop covers angle φ • Use example where φ = π (half circle) – Define direction into page as positive...
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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PHY2049_06-10-11 - 1 Phy2049 Magnetism • Last lecture...

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