lec22_022807_-_Magnetics_III_with_answers

lec22_022807_-_Magnetics_III_with_answers - Magnetics III...

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Unformatted text preview: Magnetics III Exam II: Thursday, March 1st, 7:30 9:30 p.m. Practice Exam on LON-CAPA LON Room schedule information on LON-CAPA (same as Exam I) LON AEW: tonight from 6 8 p.m. (Hill Hall 209) 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 Magnetic Force: Magnetic Torque: Fmag = qv B dFmag = Id B ^ ^ j j d r = dy ^ x i - y ^ = - xdy k ( ) B = B = NIA dB = - Magnetic Dipole Moment: 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 ^ Id r o Id r Biot-Savart Law: dB = o = Biot4 r 2 4 r 3 Wednesday, 28 February 2007 1 Wednesday, 28 February 2007 2 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 Wednesday, 28 February 2007 3 Wednesday, 28 February 2007 4 1 y B-Field of a Finite Length Wire d Question The 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 A. Zero I2 For a = b: I B=- o 2x a a2 + x2 For Infinite wire: a , b - a 2 , - b - 2 II B. 2 o 1 2 down 2d o I1 I 2 down C. 1 2d o I1 I 2 up 2d II E. 1 o 1 2 up 2d D. 2 5 Wednesday, 28 February 2007 d I1 F1 = I1 1 B2 B2 = o I 2 2d B2 direction? 1 F1 = I1 1 B2 = Force per unit length = B=- Wednesday, 28 February 2007 o I ^ I ^ k =- o k sin + sin 4x 2 2 2x o I1 I 2 2d = 1 F1 o I1 I 2 2d 6 Question The force that Wire 2 will feel due to Wire 1 is: A. Zero I2 d I1 F2 = I 2 2 Question The force that Wire 1 will feel due to Wire 2 is: B1 A. Zero II B. 2 o 1 2 down 2d o I1 I 2 down C. 1 2d o I1 I 2 up 2d II E. 1 o 1 2 up 2d D. 2 Wednesday, 28 February 2007 I B1 = o 1 2d I2 d I1 F1 = I1 1 B2 B1 direction? 2 II B. 2 o 1 2 down 2d o I1 I 2 down C. 1 2d o I1 I 2 up 2d II E. 1 o 1 2 up 2d D. 2 Wednesday, 28 February 2007 B2 = o I 2 2d B2 direction? 1 F2 = I 2 2 B1 = Force per unit length = o I1 I 2 2d = 2 F1 = I1 1 B2 = Force per unit length = o I1 I 2 2d = 1 F2 o I1 I 2 2d 7 F1 o I1 I 2 2d 8 2 Measuring Magnetic Field Measuring Magnetic Field Negative charge carriers feel a force FB = qvd B and are pushed toward the top of the metal strip E Once enough charges are piled up such that the electric force balances out the magnetic force V Va > Vb Forces are balanced: Current Density: qE z + qvd B y = 0 E z = -vd B y J x = nqvd ( Same thing happens to positive charge carriers (Va < Vb ) ) Ez = - Jx By nq nq = - J x By Ez =- J x By d Vz 10 Hall Effect (1879) Wednesday, 28 February 2007 9 Wednesday, 28 February 2007 By = -nq Vz J xd Hall Probe B = ? z y 2 mm B 1.5 cm V = -0.40 V I = 75 A x V = +0.40 V n = 11.6 10 28 m -3 By = -nq Vz J xd Charge carrier? Electrons! m ) (1.5 10 (75 )(2)10 A -2 -3 = - 11.6 10 28 m -3 - 1.6 10-19 C ( )( m 0.80 10 -6 V 1.5 10- 2 m = 0.40 T Wednesday, 28 February 2007 )( ( ) ) 11 3 ...
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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.

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