hw11 - wilson(kw22588 – HW11 – Mackie –(20224 1 This...

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Unformatted text preview: wilson (kw22588) – HW11 – Mackie – (20224) 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 (part 1 of 3) 10.0 points A 46 . 5 mA current is carried by a uniformly wound air-core solenoid with 208 turns as shown in the figure below. The permeability of free space is 1 . 25664 × 10- 6 N / A 2 . 4 6 . 5 m A 10 . 9 mm 15 . 6 cm Compute the magnetic field inside the solenoid. Correct answer: 7 . 79115 × 10- 5 T. Explanation: Let : N = 208 , ℓ = 15 . 6 cm , I = 46 . 5 mA , and μ = 1 . 25664 × 10- 6 N / A 2 . I d ℓ The magnetic field inside the solenoid is B = μ nI = μ parenleftbigg N ℓ parenrightbigg I = (1 . 25664 × 10- 6 N / A 2 ) parenleftbigg 208 . 156 m parenrightbigg × (0 . 0465 A) = 7 . 79115 × 10- 5 T . 002 (part 2 of 3) 10.0 points Compute the magnetic flux through each turn. Correct answer: 7 . 27017 × 10- 9 Wb. Explanation: Let : B = 7 . 79115 × 10- 5 T , and d = 10 . 9 mm = 0 . 0109 m . The magnetic flux through each turn is Φ = B A = B parenleftBig π 4 d 2 parenrightBig = (7 . 79115 × 10- 5 T) π 4 (0 . 0109 m) 2 = 7 . 27017 × 10- 9 Wb . 003 (part 3 of 3) 10.0 points Compute the inductance of the solenoid. Correct answer: 0 . 0325203 mH. Explanation: The inductance of the solenoid is L = N Φ I = (208) (7 . 27017 × 10- 9 Wb) . 0465 A = . 0325203 mH . 004 (part 1 of 2) 10.0 points The current in a 127 mH inductor changes with time as I = bt 2- a t . With a = 2 A / s and b = 4 A / s 2 , find the magnitude of the induced emf , |E| , at t = 1 s. Correct answer: 0 . 762 V. Explanation: wilson (kw22588) – HW11 – Mackie – (20224) 2 Let : L = 127 mH = 0 . 127 H , b = 4 A / s 2 , a = 2 A / s , and t = 1 s . From Faraday’s Law, the induced emf E is proportional to the rate of change of the magnetic flux, which in turn is proportional to the rate of change of the current. This is expressed as E = L dI dt = L d dt ( bt 2- a t ) = L (2 bt- a ) At 1 s ,the magnitude of the induced emf is |E| = (0 . 127 H) vextendsingle vextendsingle vextendsingle 2(4 A / s 2 )(1 s)- 2 A / s vextendsingle vextendsingle vextendsingle = . 762 V . 005 (part 2 of 2) 10.0 points At what time is the emf zero? Correct answer: 0 . 25 s. Explanation: From Part 1, the induced emf is zero when dI dt = 0 2 bt- a = 0 t = (2 A / s) 2 (4 A / s 2 ) = . 25 s . 006 10.0 points An inductor in the form of an air-core solenoid contains 545 turns, is of length 23 . 4 cm, and has a cross-sectional area of 1 . 2 cm 2 ....
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hw11 - wilson(kw22588 – HW11 – Mackie –(20224 1 This...

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