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25-1 - Kapoor(mk9499 homework25 Turner(60230 This print-out...

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Kapoor (mk9499) – homework25 – Turner – (60230) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An electric motor has 114 turns of wire wrapped on a rectangular coil, of dimensions 3 . 13 cm by 4 . 35 cm. Assume that the mo- tor uses 10 . 1 A of current and that a uniform 0 . 35 T magnetic field exists within the motor. What is the maximum torque delivered by the motor? Correct answer: 0 . 548691 N m. Explanation: Let : w = 3 . 13 cm , = 4 . 35 cm , I = 10 . 1 A , and B = 0 . 35 T . Force : F = I × B Torque : τ = r × F Power : P = τ ω . Magnetic force exerted on the 4 . 35 cm side is, F = N I B = (114 turns) (10 . 1 A) (0 . 0435 m) (0 . 35 T) = 17 . 5301 N . So, the maximum torque on the loop due to the magnetic field is τ max = F × w = (17 . 5301 N) × (0 . 0313 m) = 0 . 548691 N m . 002 (part 2 of 2) 10.0 points If the motor rotates at 2740 rev / min, what is the peak power produced by the motor? Correct answer: 157 . 437 W. Explanation: P = τ ω = (0 . 548691 N m) (286 . 932 s - 1 ) = 157 . 437 W . 003 10.0 points A long solenoid has a coil made of fine wire inside it and coaxial with it. I Outside solenoid has n turns per meter Inside coil has N turns r R Given a varying current I in the outer solenoid, what is the emf induced in the inner loop? 1. E = - π r 2 μ 0 n N d I dt 2. E = - π r μ 0 N d I dt 3. E = - π r 2 μ 0 n d I dt 4. E = - π R 2 μ 0 n N d I dt correct 5. E = - π r μ 0 n d I dt 6. E = - π R μ 0 N d I dt 7. E = - π R 2 μ 0 n d I dt 8. E = - π R μ 0 n N d I dt 9. E = - π r μ 0 n N d I dt 10. E = - π R μ 0 n d I dt Explanation: The magnetic field of a solenoid is B = μ 0 n I . The magnetic flux is Φ B = B · A = ( μ 0 n I ) ( π R 2 ) ,
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Kapoor (mk9499) – homework25 – Turner – (60230) 2 so the emf is E = - d Φ B dt = - π R 2 μ 0 n N d I dt . We are interested in the emf in the inner coil, so we use the smaller area of the inner coil rather than the larger solenoid area. keywords: 004 (part 1 of 2) 10.0 points Two infinitely long solenoids (seen in cross section in the figure below) thread a circuit. Given: a = 0 . 9 m , r 1 = 0 . 2 m , r 2 = 0 . 1 m , Δ B Δ t = 160 T / s , R l = 9 Ω , R m = 4 Ω , and R r = 2 . 6 Ω , as in the figure below. The magnitude of B inside each solenoid is the same and is 300 T at time t = 0. B in B out I l I m I r r 2 R m R l R r r 1 a a a What is the magnitude of the current, I m , in middle resistor, R m ? Correct answer: 1 . 39706 A. Explanation: Basic Concept: Faraday’s Law: E = - d Φ B dt Ohm’s Law: I = V R Junction Rule: n i =1 I i = 0 Solution: Note: The side-length, a , of the circuit loop is not necessary for this problem. Neither is the magnitude of B at time t = 0. From Faraday’s law, the induced emf in the left loop (Loop 1) is |E 1 | = d Φ B dt = A 1 d B dt = π r 2 1 d B dt = 20 . 1062 V and the direction of this emf is counter- clockwise. Similarly, the induced emf in the right loop (Loop 2) is |E 2 | = d Φ B dt = A 2 d B dt = π r 2 2 d B dt = 5 . 02655 V and the direction of this emf is clockwise.
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