# OMT 03 - terry (ect328) oldmidterm 03 Turner (59130) 1 This...

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Unformatted text preview: terry (ect328) oldmidterm 03 Turner (59130) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A copper strip (8 . 47 10 22 electrons per cu- bic centimeter) 11 . 5 cm wide and 0 . 19 cm thick is used to measure the magnitudes of unknown magnetic fields that are perpendic- ular to the strip. The charge on the electron is 1 . 6 10 19 C. Find the magnitude of B when the current is 24 A and the Hall voltage is 1 . 2 V. Correct answer: 1 . 28744 T. Explanation: Let : n = 8 . 47 10 22 cm 3 , = 8 . 47 10 28 m 3 , q = 1 . 6 10 19 C , t = 0 . 19 cm = 0 . 0019 m , w = 11 . 5 cm = 0 . 115 m , I = 24 A , and V H = 1 . 2 V = 1 . 2 10 6 V . The current in the metal strip is I = nq v d A = nq v d ( w t ) v d w = I nq t The Hall voltage is V H = v d w B B = V H v d w B = nq tV H I = (8 . 47 10 28 m 3 ) (1 . 6 10 19 C) 24 A (0 . 0019 m) (1 . 2 10 6 V) = 1 . 28744 T . 002 10.0 points A small rectangular coil composed of 57 turns of wire has an area of 41 cm 2 and carries a current of 1 . 1 A. When the plane of the coil makes an angle of 26 with a uniform magnetic field, the torque on the coil is 0 . 14 N m. What is the magnitude of the magnetic field? Correct answer: 0 . 605922 T. Explanation: Let : N = 57 turns , I = 1 . 1 A , = 26 , A = 41 cm 2 = 0 . 0041 m 2 , and = 0 . 14 N m . The magnetic force on the current is vector F = I vector vector B and the torque is vector = vector r vector F , so the torque on the loop due to the magnetic field is = 2 F r cos = ( N I B ) w cos = N I B ( w ) cos = N I B A cos , where A is the area of the loop and is the angle between the plane of the loop and the magnetic field. The magnetic field from above is B = N I A cos = . 14 N m (57 turns) (1 . 1 A) (0 . 0041 m 2 ) cos(26 ) = . 605922 T . 003 10.0 points Given: Assume the bar and rails have neg- ligible resistance and friction. terry (ect328) oldmidterm 03 Turner (59130) 2 In the arrangement shown in the figure, the resistor is 5 and a 5 T magnetic field is directed out of the paper. The separation between the rails is 4 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 9 m / s . m 1g 9 m / s 5 5 T 5 T I 4m At what rate is energy dissipated in the resistor? Correct answer: 6480 W. Explanation: Basic Concept: Motional E E = B v . Ohms Law I = V R . Solution: The motional E induced in the circuit is E = B v = (5 T) (4 m) (9 m / s) = 180 V . From Ohms law, the current flowing through the resistor is I = E R = B v R = (5 T) (4 m) (9 m / s) R = 36 A ....
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## This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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OMT 03 - terry (ect328) oldmidterm 03 Turner (59130) 1 This...

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