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oldhw 26 - nguyen(jmn727 oldhomework 26 Turner(59070 This...

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nguyen (jmn727) – oldhomework 26 – Turner – (59070) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In an RL series circuit, an inductor of 3 . 2 H and a resistor of 7 . 08 Ω are connected to a 24 . 6 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Correct answer: 19 . 3163 J. Explanation: Let : L = 3 . 2 H , R = 7 . 08 Ω , and E = 24 . 6 V . The current in an RL circuit is I = E R parenleftBig 1 - e Rt/L parenrightBig . The final equilibrium value of the current, which occurs as t → ∞ , is I 0 = E R = 24 . 6 V 7 . 08 Ω = 3 . 47458 A . The energy stored in the inductor carrying a current 3 . 47458 A is U = 1 2 L I 2 = 1 2 (3 . 2 H) (3 . 47458 A) 2 = 19 . 3163 J . 002 (part 1 of 3) 10.0 points A long solenoid carries a current I 2 . Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure below. 2 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 The current I 2 is held constant. The energy stored in the solenoid is given by 1. U = μ 0 2 N 2 A 2 2 I 2 2 2. U = μ 0 2 N 2 2 A 2 2 I 2 2 correct 3. U = μ 0 2 N 2 2 2 A 2 I 2 2 4. U = μ 0 2 N 2 2 A 2 2 I 2 5. U = μ 0 2 A 2 2 I 2 6. U = μ 0 2 N 2 2 A 2 I 2 2 7. U = 1 2 μ 0 N 2 2 A 2 I 2 2 8. U = μ 0 2 N 2 2 A 1 2 I 2 2 Explanation: The magnetic energy density is given by B 2 2 μ 0 . Inside the solenoid the magnetic field is B = μ 0 N 2 I 2 2 , and the volume enclosed by the solenoid is A 2 2 , so U = 1 2 μ 0 parenleftbigg μ 0 N 2 2 I 2 parenrightbigg 2 A 2 2 = μ 0 2 N 2 2 A 2 2 I 2 2 . 003 (part 2 of 3) 10.0 points The mutual inductance M 12 between the coil and the solenoid is given by
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nguyen (jmn727) – oldhomework 26 – Turner – (59070) 2 1. M 12 = μ 0 N 1 N 2 2 2. M 12 = μ 0 N 1 N 2 A 2 2 correct 3. M 12 = μ 0 N 1 N 2 A 1 4. M 12 = μ 0 N 1 N 2 A 1 2 5. M 12 = μ 0 N 2 A 2 2 6. M 12 = μ 0 N 1 N 1 N 2 7. M 12 = μ 0 2 N 1 N 2 A 2 Explanation: The mutual inductance M 12 of loop 1 with respect to loop 2 is defined as M 12 N 1 Φ 12 I 2 , where Φ 12 is the flux through a single loop 1 due to loop 2. Since the magnetic field inside the coil is restricted to the part inside the solenoid, we have Φ 12 = μ 0 N 2 2 I 2 A 2 , so M 12 = μ 0 N 2 N 1 A 2 2 .
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