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Ch19-h1-solutions

# Ch19-h1-solutions - liu(ql744 Ch19-h1 chiu(56565 This...

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This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Why does the the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery? A. Very little energy is dissipated in the thick connecting wires. B. The electric field in the connecting wires is very small, so emf E bulb L bulb . C. The electric field in the connecting wires is zero, so emf = E bulb L bulb . D. The current in the connecting wires is smaller than the current in the bulb. E. All the current is used up in the bulb, so the connecting wires don’t matter. Correct answer: A,B. Explanation: (A) and (B) are correct. For (B), when there are two connecting wires, applying the Loop Rule gives emf = Δ V 1 + Δ V bulb + Δ V 2 The potential di ff erence across a connect- ing wire is Δ V 1 = E 1 L 1 . Because the cross sectional area of the connecting wires is much larger than the cross sectional area of the fil- ament, then for each wire E 1 E bulb . Thus, Δ V 1 Δ V bulb . For (A), we need to write down the above equation in terms of the energy equation, Δ U = q Δ V . Thus, q (emf) = q Δ V 1 + q Δ V bulb + q Δ V 2 Since Δ V 1 Δ V bulb , then the energy loss of electrons through the connecting wire is much smaller than through the bulb and q Δ V 1 is negligible. As a result, emf Δ V bulb . C) is incorrect because E=0 necessarily im- plies I=0. (D) is incorrect because it violates charge conservation. (E) is incorrect because current is not ”used up.” 002 (part 1 of 5) 10.0 points The following questions correspond to the fig- ure shown, consisting of two flashlight bat- teries and two Nichrome wires of di ff erent lengths and di ff erent thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires). + - + - The thin wire is 54 cm long, and its diame- ter is 0 . 2 mm. The thick wire is 23 cm long, and its diameter is 0 . 37 mm. The emf of each flashlight battery is 1 . 5 V. Determine the steady-state electric field in- side each Nichrome wire. Remember that in the steady state you must satisfy both the current node rule and energy conservation. These two principles give you two equations for the two unknown fields. Find the electric field in the thin wire first. Correct answer: 4 . 94069 V / m. Explanation: Apply the loop rule. Call the thin wire 1 and the thick wire 2. 2 emf - E 1 L 1 - E 2 L 2 = 0 Apply the node rule. i 1 = i 2 nA 1 uE 1 = nA 2 uE 2 A 1 E 1 = A 2 E 2 E 1 = A 2 A 1 E 2 = (0 . 37 mm) 2 (0 . 2 mm) 2 E 2 = 3 . 4225 E 2 Substitute into the loop equation.

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2 emf - 3 . 4225 E 2 L 1 - E 2 L 2 = 0 2 emf - E 2 (3 . 4225 L 1 + L 2 ) = 0 E 2 = 2 emf 3 . 4225 L 1 + L 2 = 2(1 . 5 V) (3 . 4225)(0 . 54 m) + (0 . 23 m) = 1 . 44359 V / m So, E 1 = 3 . 4225 E 2 = 4 . 94069 V / m . 003 (part 2 of 5) 10.0 points Find the electric field in the thick wire. Correct answer: 1 . 44359 V / m.
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