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Forces on wire How to Whiteboards

Forces on wire How to Whiteboards. N. I. S. Demo - Wire loop with magnet What about the direction???. Force on a current carrying wire: F = IlxB = IlBsin  (rt hand direction) x is vector cross product F = force on wire (N) I = current (A) l = length of wire in B field (m)

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Forces on wire How to Whiteboards

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  1. Forces on wire • How to • Whiteboards

  2. N I S Demo - Wire loop with magnet What about the direction??? • Force on a current carrying wire: • F = IlxB = IlBsin (rt hand direction) • x is vector cross product • F = force on wire (N) • I = current (A) • l = length of wire in B field (m) • B = magnetic field in Teslas (1 T = 1 N/Am) • 1 T = 10,000 Gauss •  = Angle twixt B and l (tail to tail)

  3. Force on a current carrying wire: • F = ILxB = IlBsin • F = force on wire (N) • I = current (A) • l = length of wire in B field (m) • B = magnetic field in Teslas (1 T = 1 N/Am) •  = Angle twixt B and l (tail to tail) • Direction of force: • Index finger - Direction of Il • Middle finger - Direction of B • Thumb - Direction of Force • Review demo

  4. Whiteboards - Direction

  5. Which way is the force? I B outa the page

  6. Which way is the force? I B inta da page

  7. Which way is the force? B I outa da page

  8. Which way is the force? B I inta da page

  9. Which way is the force? B I ??????

  10. Which way is the force? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I B down

  11. Which way is the force? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I B Show other way left

  12. Which way is the force? x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x I B down

  13. Which way is the force? x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x I B down to the left

  14. Which way is the magnetic field? I F inta da page

  15. Which way is the current? B F outa da page

  16. Which way is the force? I N S inta da page

  17. Which way is the force? N I S outa da page

  18. Which way is the force? N S outa da page

  19. F = ? Il = Up B = W (S) • F = ? Il = N B = E (D) • F = Up Il = W B = ? (S) • F = N Il = Down B = ? (W) • F = E Il = ? B = N (D) • F = W Il = ? B = Up (S) N W E S

  20. Whiteboards - Force: F = IlBsin 1 | 2 | 3 | 4 | 5

  21. A 0.15 T magnetic field is 27o east of North What’s the force on a 3.2 m long wire if the current is 5.0 A to the West?  = 90o + 27o = 117o F = IlBsin F = (5.0 A)(3.2 m)(0.15 T)sin(117o) = 2.13841 N E x U = S N W E S 2.1 N vertically downward

  22. What current in what direction would you need to have a force of 10.0 N to the west in 50.0 cm of wire perpendicular to Earth’s magnetic field of 0.5 x 10-4 T to the North? F = IlBsin 10.0 N = I(.500 m)(.5 x 10-4 T)sin(90o) I = 4 x 105 A ? x N = S (U) N W E S 4 x 105 A upward

  23. A 17 cm wire forms a 37o angle with an unknown magnetic field. What is the magnetic field if the force equals .015 N and I = 5.0 A? F = IlBsin .015 N = (5.0 A)(.170 m)Bsin(37o) B = 2.9 x 10-2 T 2.9 x 10-2 T

  24. Find I and its direction in the B-Field Balanced Pulley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B = .25 T 1.5 m 3.2 N Which way is the force? F = IlBsin 3.2 N = I(1.5 m)(.25 T)sin(90o) I = 8.5 A, ACW, down on left side 8.5 A, ACW

  25. Find F and its direction x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x B = .15 T 20. cm 12v 6.0  F = IlBsin, I = V/R = 2.0 A F = (2.0 A)(.20 m)(.15 T)sin(90o) F = .060 N, Up .060 N, Up

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