lec21_022607_-_Magnetics_II_with_answers

lec21_022607_-_Magnetics_II_with_answers - Magnetics II...

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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 LON-CAPA LON Room schedule information on LON-CAPA (same as Exam I) LON- Magnetic Force on a Curved I-carrying 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 current-carrying 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 B-Field B- Current Carrying Loop in an Uniform B-Field 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 E-Field and B-Field 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 "Biot-Savart 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 B-Field 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 B-Field of a Finite Length Wire d y B-Field 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 B-Field 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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