1 Magnetic Force on a Charged Particle

2 Chapter Assignments Chapter 20 HW#1 Q2,3,9-11 and P1,7,8, HW#2 P11(c), 12-14,17(c), 26,29,30(c), 48(c) and 49(c) Chapter 21 HW#3 Q2,4,6,8 and P1-5,9(c),10(c),14,15

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Force on Electric Current in Magnetic Field

14 F = qvB = (3.0x10 -6 C)(3x10 5 m/s)(200x10 -3 T) F = 1.8x10 -1 N down No force, vectors are parallel

15 F = qVB sin  direction: up

16 Motion of Charged Particles in a Magnetic Field

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19 +q is forced down toward the bottom of the page

20 F = qvB B = F/qv = 2000N C(1.6x10 -19C)(3x105 m/s) B = _____T

21 F = qVB = (1.60x C) (300,000 m/s) (2000x10 -3 T) Direction: out of the page

22 F = Eq = 2000 N/c (1.60 x c) = Direction: out of page (use left hand for electron)

23 F = qVB = 1.60 x c (300,000 m/s)(2 T) = Direction: down

24 Magnetic Force on Current Carrying Wires

25 B = F/qV = 400V/20,000 m/s = A volt is work done per unit charge (J/q) Direction: into the page

26 F = I L B sin  F = (30A)(100m)(0.4T)(sin90) F = _____________N

27 F = I L B  B = F / I L = (40N/m) / 15A = Current moving to the right

28 Sources of Magnetic Fields

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30 Hand rules When current flows through a wire, it generates a magnetic field around the wire. The direction of the field is determined with one of the hand rules

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37 I 1 = 10.0 A I 2 = 13.0 A 7.0 m r2r2 r 1 Opposite Directions could be zero

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40 How do we determine which way is north? When current is flowing through a coil, the coil acts like a bar magnet. We can predict which way is north using another rule

41  B A B - out A - in

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43 Magnetic Flux

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48 a)0 b)Max:  = BAcos  = 4.2x10 -3 T(6m 2 )cos  = ___ Weber c  = BAcos  = 4.2x10 -3 T(6m 2 )cos30 = ___ Wb

49 F = B  r 2 cos 40 r = sr [  / B  cos 40] r = sr [.250Wb / (0.050T)(3.14)(cos40) = ______m

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51 astr.gsu.edu/hbase/electric/farlaw.htmlhttp://hyperphysics.phy- astr.gsu.edu/hbase/electric/farlaw.html

52 E = (500 x T / 0.01 s) = ___V

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54 sics/Lenz's.htmlhttp:// sics/Lenz's.html

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57 clockwise

58 counterclockwise

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61 B =  o I / 2  r  I = 2  Br /  o =  = 2  x Power = I 2 r =

62 Direction of current through 3 ohm resistor is down

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