Hw10 solutions - homework 10 – SMITH TAYLOR – Due 4:00 am 1 Question 1 chap 31 sect 4 part 1 of 1 10 points A long solenoid has a coil made of

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Unformatted text preview: homework 10 – SMITH, TAYLOR – Due: Mar 31 2008, 4:00 am 1 Question 1, chap 31, sect 4. part 1 of 1 10 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 μ nN dI dt 2. E = − π Rμ n dI dt 3. E = − π Rμ N dI dt 4. E = − π r μ n dI dt 5. E = − π R 2 μ n dI dt 6. E = − π R 2 μ nN dI dt correct 7. E = − π r 2 μ n dI dt 8. E = − π Rμ nN dI dt 9. E = − π r μ N dI dt 10. E = − π r 2 μ nN dI dt Explanation: The magnetic field of a solenoid is B = μ nI . The magnetic flux is Φ B = B · A = ( μ nI ) ( π R 2 ) , so the emf is E = − d Φ B dt = − π R 2 μ nN dI 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. Question 2, chap 31, sect 2. part 1 of 1 10 points A coil with N 1 = 11 . 8 turns and radius r 1 = 8 . 98 cm surrounds a long solenoid of radius r 2 = 1 . 44 cm and N 2 ℓ 2 = n 2 = 1200 m − 1 (see the figure below). The current in the solenoid changes as I = I sin( ω t ), where I = 5 A and ω = 120 rad / s. R a b E ℓ 2 ℓ 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 What is the magnitude of the induced emf , E AB , across the 11 . 8 turn coil at t = 1000 s? Correct answer: 0 . 00579692 V (tolerance ± 1 %). Explanation: Faraday’s Law for solenoid E = − N d Φ dt = − N d ( BA ) dt . Magnetic field induced by solenoid B = μ nI . Faraday’s Law for solenoid E = − N d Φ dt = − N d ( BA ) dt . homework 10 – SMITH, TAYLOR – Due: Mar 31 2008, 4:00 am 2 Magnetic field induced by inner solenoid B = μ nI . So the induced emf is E = − N 1 d Φ dt = − N 1 d ( B A 2 ) dt = − N 1 A 2 dB dt = − N 1 A 2 d ( μ n 2 I ) dt = − μ N 1 n 2 A 2 dI dt = . 00579692 V . Question 3, chap 31, sect 6. part 1 of 3 10 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. 198 mH 7 . 47 Ω 6 . 3 V S b a If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 113 mA? Correct answer: 3 . 81299 ms (tolerance ± 1 %). Explanation: Let : R = 7 . 47 Ω , L = 198 mH , and E = 6 . 3 V . L R E S b a The time constant of an RL circuit is τ = L R = . 198 H 7 . 47 Ω = 0 . 026506 s . The final current reached in the circuit is I = E R = 6 . 3 V 7 . 47 Ω = 0 . 843373 A . The switch is in position a in an RL circuit connected to a battery at t = 0 when I = 0....
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This note was uploaded on 04/10/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Hw10 solutions - homework 10 – SMITH TAYLOR – Due 4:00 am 1 Question 1 chap 31 sect 4 part 1 of 1 10 points A long solenoid has a coil made of

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