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 , . 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 . 8 μ V = 1 . 8 × 10 - 6 V . The current in the metal strip is I = nq v d A = d ( wt ) v d w = I nq t The Hall voltage is V H = v d wB 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 . 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 . 2A , θ = 36 , A = 41 cm 2 . 0041 m 2 , and τ . 05 N m . The magnetic Force on the current is ' F = I ' & × ' B and the torque is = ' r × ' F , so the torque on the loop due to the magnetic feld is τ =2 Fr cos θ =( N I & B ) w cos θ = N I B ( & 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 ± 1g 5m / s 7T I At what rate is energy dissipated in the resistor? Correct answer: 5104 . 17 W. Explanation: Basic Concept: Motional E E = B & v . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B & 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 & v R = (7 T) (5 m) (5 m / s) R = 29 . 1667 A . The power dissipated in the resistor is P = I 2 R = B 2 & 2 v 2 R 2 R = B 2 & 2 v 2 R = (7 T) 2 (5 m) 2 (5 m / s) 2 (6 Ω) = 5104 . 17 W . Note: First of four versions. 004 10.0 points A coil is wrapped with 582 turns of wire on the perimeter of a circular frame (of radius 37 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic ²eld is directed perpendicular to the plane of the coil. This ²eld changes at a constant rate from 22 mT to 64 mT in 54 ms.
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Exam - Kapoor (mk9499) oldmidterm 03 Turner (60230) This...

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