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Unformatted text preview: nguyen (jmn727) – oldmidterm 03 – Turner – (59070) 1 This printout should have 19 questions. Multiplechoice 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) 13 . 4 cm wide and 0 . 15 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 23 A and the Hall voltage is 2 μ V. Correct answer: 1 . 76765 T. Explanation: Let : n = 8 . 47 × 10 22 cm − 3 , = 8 . 47 × 10 28 m − 3 , q = 1 . 6 × 10 − 19 C , t = 0 . 15 cm = 0 . 0015 m , w = 13 . 4 cm = 0 . 134 m , I = 23 A , and V H = 2 μ V = 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) 23 A × (0 . 0015 m) (2 × 10 − 6 V) = 1 . 76765 T . 002 10.0 points A small rectangular coil composed of 75 turns of wire has an area of 25 cm 2 and carries a current of 1 . 6 A. When the plane of the coil makes an angle of 38 ◦ with a uniform magnetic field, the torque on the coil is 0 . 06 N m. What is the magnitude of the magnetic field? Correct answer: 0 . 253804 T. Explanation: Let : N = 75 turns , I = 1 . 6 A , θ = 38 ◦ , A = 25 cm 2 = 0 . 0025 m 2 , and τ = 0 . 06 N m . The magnetic force on the current is vector F = I vector ℓ × vector B and the torque is vector τ = vector r × vector F , so the torque on the loop due to the magnetic field 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 field. The magnetic field from above is B = τ N I A cos θ = . 06 N m (75 turns) (1 . 6 A) (0 . 0025 m 2 ) cos(38 ◦ ) = . 253804 T . 003 10.0 points Given: Assume the bar and rails have neg ligible resistance and friction. nguyen (jmn727) – oldmidterm 03 – Turner – (59070) 2 In the arrangement shown in the figure, the resistor is 9 Ω and a 4 T magnetic field is directed out of the paper. The separation between the rails is 2 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 7 m / s . m ≪ 1g 7 m / s 9Ω 4 T 4 T I 2m At what rate is energy dissipated in the resistor? Correct answer: 348 . 444 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 = (4 T) (2 m) (7 m / s) = 56 V . From Ohm’s law, the current flowing through the resistor is I = E R = B ℓ v R = (4 T) (2 m) (7 m / s) R = 6 . 22222 A ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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