Exam - Kapoor (mk9499) oldmidterm 03 Turner (60230) This...

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Kapoor (mk9499) – oldmidterm 03 – Turner – (60230) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A copper strip (8 . 47 × 10 22 electrons per cu- bic centimeter) 7 . 4 cm wide and 0 . 02 cm thick is used to measure the magnitudes oF un- known magnetic felds that are perpendicular to the strip. The charge on the electron is 1 . 6 × 10 19 C. ±ind the magnitude oF B when the current is 24 A and the Hall voltage is 1 . 8 μ V. Correct answer: 0 . 20328 T. Explanation: Let : n = 8 . 47 × 10 22 cm 3 , = 8 . 47 × 10 28 m 3 , q = 1 . 6 × 10 19 C , t = 0 . 02 cm = 0 . 0002 m , w = 7 . 4 cm = 0 . 074 m , I = 24 A , and V H = 1 . 8 μ V = 1 . 8 × 10 6 V . The current in the metal strip is I = n q v d A = n q v d ( w t ) v d w = I n q t The Hall voltage is V H = v d w B B = V H v d w B = n q tV H I = (8 . 47 × 10 28 m 3 ) (1 . 6 × 10 19 C) 24 A × (0 . 0002 m) (1 . 8 × 10 6 V) = 0 . 20328 T . 002 10.0 points A small rectangular coil composed oF 49 turns oF wire has an area oF 41 cm 2 and carries a current oF 1 . 2 A. When the plane oF the coil makes an angle oF 36 with a uniForm magnetic feld, the torque on the coil is 0 . 05 N m. What is the magnitude oF the magnetic feld? Correct answer: 0 . 256361 T. Explanation: Let : N = 49 turns , I = 1 . 2 A , θ = 36 , A = 41 cm 2 = 0 . 0041 m 2 , and τ = 0 . 05 N m . The magnetic Force on the current is ± F = I ± × ± B and the torque is ± τ = ± r × ± F , so the torque on the loop due to the magnetic feld is τ = 2 F r cos θ = ( N I ℓ B ) w cos θ = N I B ( ℓ w ) cos θ = N I B A cos θ , where A is the area oF the loop and θ is the angle between the plane oF the loop and the magnetic feld. The magnetic feld From above is B = τ N I A cos θ = 0 . 05 N m (49 turns) (1 . 2 A) (0 . 0041 m 2 ) cos(36 ) = 0 . 256361 T . 003 10.0 points Given: Assume the bar and rails have neg- ligible resistance and Friction. In the arrangement shown in the fgure, the resistor is 6 Ω and a 7 T magnetic feld is directed into the paper. The separation
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Kapoor (mk9499) – oldmidterm 03 – Turner – (60230) 2 between the rails is 5 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 5 m / s . m 1 g 5 m / s 6 Ω 7 T 7 T I 5 m At what rate is energy dissipated in the resistor? Correct answer: 5104 . 17 W. Explanation: Basic Concept: Motional E E = B ℓ v . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B ℓ v = (7 T) (5 m) (5 m / s) = 175 V . From Ohm’s law, the current ±owing through the resistor is I = E R = B ℓ v R = (7 T) (5 m) (5 m / s) R = 29 . 1667 A .
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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Exam - Kapoor (mk9499) oldmidterm 03 Turner (60230) This...

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