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Unformatted text preview: Magnetics II Exam II: Thursday, March 1st, 7:30 9:30 p.m. Review recitations this week (NO QUIZ) Practice Exam on LONCAPA LON Room schedule information on LONCAPA (same as Exam I) LON Magnetic Force on a Curved Icarrying Wire
F = qv B
y I
d dF dF
y B
d dq
x dF = (dq )v B d = dq B dt dq = d B dt dF = Id B v= d dt dF = Id B Magnetic Force On a moving charge: B Fmag = qv B d On a currentcarrying wire: current Fmag = I B
dFmag = Id B R d x Motion of a charge in an Uniform magnetic field qB Circular motion: = m qB Cyclotron frequency: R F= f = Monday, 26 February 2007 2m d = Rd  sin i^ + cos ^ j
1 Monday, 26 February 2007 ( ) )( ^ =  IRBd (cos i + sin ^ ) j ^ j dF =  IRB (cos i + sin ^ ) d ^ =  IRB[sin i  cos ^ ] j =  IRB[(0) i^  ( 1  1) ^ ] j ( 0 ^ ^ = IRd  sin i + cos ^  B k j ) 0 = 2 IRB ^ j 2 An electron enters a capacitor as shown below. If you do not want the electron to be deflected, which direction should the magnetic field be directed?
v
 Velocity Selector Bms Mass Spectrometer
v F = qv B E
 Path
 + + + + + + + A. B. C. D. E. Up Down e To the right Into the page Out of the page F
 R + + + + + + + + + + + + + + F = qvBms = m
E Bcap Detector v2 R   Path
v FE = qE FE FB
qE = qvB v= m=
= B qBms R v qBms Bcap
E R F = ma = 0 FE  FB = 0 FE = FB v= E B
3 Monday, 26 February 2007 4 Monday, 26 February 2007 1 Current Carrying Loop in an Uniform BField B Current Carrying Loop in an Uniform BField B is maximum when the normal of the loop is to B ! = IAB sin Monday, 26 February 2007 5 Monday, 26 February 2007 is 0 when the normal of the loop is to B , and it is in stable equilibrium! is still 0 when the normal of the loop is anti to B , but it is antiin unstable equilibrium! 6 Comparing EField and BField EB Source of Magnetic Field
Moving charges experience a magnetic force:
F = qv B dF = Id B Electric Dipole Moment: Magnetic Dipole Moment: Reasonable to assume that it is moving charges that create it: p = qd = IA B = B
U B =  B
7 dB = ^ o Id r o Id r = 2 4 r 4 r 3 E = p E
UE =  p E
Monday, 26 February 2007 7 7 2 Permeability of free space: o = 4 10 T m/A = 4 10 N/A "BiotSavart Law" BiotLaw"
Monday, 26 February 2007 8 2 Question
Given the current carrying wire shown below, the magnetic field at point P is directed: A. B. C. D. E. To the right To the left Into the page Out of the page None of the above
a b y BField of a Finite Length Wire
d I r P dB =
x o Id r 4 r 3 j d = dy ^ B ^ r = xi  y ^ j r = x2 + y 2 B ^ ^ j j d r = dy ^ x i  y ^ =  xdy k ( ) dB =  o I ( xdy ) ^ k 4 (x 2 + y 2 )3 / 2
dy Ix ^ B = dB =  o k 3/ 2 2 4 b (x + y 2 )
a Monday, 26 February 2007 9 Monday, 26 February 2007 10 y BField of a Finite Length Wire
d y BField of a Finite Length Wire
d a b I r P dy Ix ^ B = dB =  o k 3/ 2 2 4 b (x + y 2 ) x
2 2 a a b I r P B=
x o I ^ k cos d 4x 
b a B
r x B
r x y x 1 cos r= x +y = = cos r x x d y x tan = y dy = x sec 2 d = tan = cos 2 x
a a = =
= o I sin 4x a
b ^ k y o I ^ [sin a + sin b ] k 4x
o I 4x
a a +x
2 2 Ix ^ xd cos 3 I ^ B= o k =  o k cos d 2 3 4  cos x 4x 
b b + b b + x2
2 ^ k Monday, 26 February 2007 11 Monday, 26 February 2007 12 3 y BField of a Finite Length Wire
d Question
The net force that Wire 1 will feel due to Wire 2 is:
^ k a b I r P x B= B o I 4x
^ k a a +x
2 2 + b b +x
2 2 For a = b: B= o I 2x a a2 + x2 For Infinite wire: a , b  a 2 ,  b  2 A. B. C. D. E. Upward Downward To the left To the right Is Zero dF1 = I1d 1 B2 B2 B=
Monday, 26 February 2007 o I ^ I ^ k = o k sin + sin 4x 2 2 2x
13 Monday, 26 February 2007 14 Question
The force that Wire 2 will feel due to Wire 1 is: A. Zero
I2
d
B1 = B1 o I1 I 2 down 2 2d o I1 down C. 1 2d
B. I1
F2 = I 2
2 o I1 2d B1 direction?
2 o I1 I 2 up 2d I E. 1 o 1 up 2d
D.
2
Monday, 26 February 2007 F2 = I 2 2 B1 =
Force per unit length = o I1 I 2 2d
=
2 F2 o I1 I 2 2d
15 4 ...
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 Fall '07
 JapGuy

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