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Unformatted text preview: d Ed =  dt ( B dA) Faraday's Law III Faraday' Relates a timevarying Bfield at some location in timeBspace to a curly Efield around that area. E d B emf =  dt B Example
What is the induced emf in the ring? = d B dt dB Fingers curl in E NC dt Quantitatively, we related integral of the curly Efield around Ea closed path to the rate of change of the magnetic flux on a surface enclosed by that path. path.
Direction: Thumb points in  R B = B dA
= B A = BA cos = BA cos(t ) emf = E d
N turns in the coil:
Friday, 23 March 2007 B = B dA
E d = N d Ed =  dt
1 Friday, 23 March 2007 d B emf =  N dt d dt ( B dA) ( B dA) = d [BA cos(t )] dt =  BA sin (t ) = BA sin (t )
2 B Example
What is the induced emf in the coil? Lenz's Law Lenz' = Ed =  = BA sin (t ) d dt ( B dA) =  ddt B R If the ring has a resistance of 2 , what is the induced current? current? Change is Bad! = IR I= = d Ed =  B dA dt BA sin (t )
R R ( ) Friday, 23 March 2007 3 Friday, 23 March 2007 4 1 Question
What will be the direction of the current through R2 immediately after the switch is closed? Question
What will be the direction of the current through R2? A. Left B. Right C. Zero D. Cannot be determined
Friday, 23 March 2007 5 A. Left B. Right C. Zero D. Cannot be determined
Friday, 23 March 2007 6 Question
What will be the direction of the current through R2 immediately after the switch opened? Application of Faraday's Law Faraday' A. Left B. Right C. Zero D. Cannot be determined
Friday, 23 March 2007 7 Friday, 23 March 2007 8 2 The bar shown is given an initial velocity V, and slides on frictionless rails. Describe its subsequent motion: Question Motional emf
Lenz's Law Faraday's Law Lenz' Faraday' d B = Binduced dt B = B dA = B dA = B dA A. It continues to move with constant velocity V; B. It accelerates to the right constantly increasing speed; C. It accelerates to the right until it reaches a constant maximum speed; D. It accelerates to the left until it stops, then reverses direction and moves to the left; E. It experiences simple harmonic motion.
Friday, 23 March 2007 9 d B d = (B x ) dt dt dx =B dt =B v = d B =B v dt = BA =B x
Friday, 23 March 2007 I= R = B v R 10 B A Single Bar
qv B
E v B A Rotating Bar
v
r dr FB = FE qvB = qE vB = E qE = Ed
= E d =E d =E
Friday, 23 March 2007 E= vB = Induced emf in an airplane: B = 0.0005 T v = 340 m/s (Mach 1) = 10 m = vB = 1.7 V!
11 A conducting bar of length rotates with a constant angular speed about a pivot at one end. A uniform Bfield is directed Bperpendicular to the plane of rotation as shown. Find the motional emf induced between the ends of the bar.
Friday, 23 March 2007 12 3 Question getting something for nothing
Suppose you have this brilliant idea: A Ferris wheel has radial metallic spokes between the hub and the circular rim (~ 10 m). These spokes move in the (~ m). magnetic field of the Earth (0.5 e4 T), so each spoke acts like (0.5 e T), a bar in the previous example. You plan to use the emf generated by the rotation of the Ferris wheel to power the lightbulbs on the wheel. Suppose the period of rotation for lightthe Ferris wheel is 1 minute. Will this idea work? minute. A. Yes B. No C. I don't know don'
Friday, 23 March 2007 13 Friday, 23 March 2007 Question follow up 14 4 ...
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This note was uploaded on 04/08/2008 for the course PHGN 200 taught by Professor Japguy during the Fall '07 term at Mines.
 Fall '07
 JapGuy

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