Ch19-h1-solutions

Ch19-h1-solutions - pan (zp695) Ch19-h1 chiu (56565) This...

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pan (zp695) – Ch19-h1 – chiu – (56565) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd 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 feld in the connecting wires is very small, so emf E bulb L bulb . C. The electric feld 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. ±or (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 f erence across a connect- ing wire is Δ V 1 = E 1 L 1 .B e c a u s et h ec r o s s sectional area oF the connecting wires is much larger than the cross sectional area oF the fl- ament, then For each wire E 1 ± E bulb .Thu s , Δ V 1 ± Δ V bulb . ±or (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 ,thentheenergyloss 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 fg- ure shown, consisting oF two ²ashlight bat- teries and two Nichrome wires oF di f erent lengths and di f erent thicknesses as shown (corresponding roughly to your own thick and thin Nichrome wires). + - The thin wire is 51 cm long, and its diame- ter is 0 . 19 mm. The thick wire is 23 cm long, and its diameter is 0 . 38 mm. The emF oF each ²ashlight battery is 1 . 8V . Determine the steady-state electric feld 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 felds. ±ind the electric feld in the thin wire frst. Correct answer: 6 . 34361 V / m. Explanation: Apply the loop rule. Call the thin wire 1 and the thick wire 2. 2emF - 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 . 38 mm) 2 (0 . 19 mm) 2 E 2 =4 E 2 Substitute into the loop equation.
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pan (zp695) – Ch19-h1 – chiu – (56565) 2 2emf - 4 E 2 L 1 - E 2 L 2 =0 - E 2 (4 L 1 + L 2 )=0 E 2 = 4 L 1 + L 2 = 2(1 . 8V) (4)(0 . 51 m) + (0 . 23 m) =1 . 5859 V / m So, E 1 =4 E 2 = 6 . 34361 V / m . 003 (part 2 of 5) 10.0 points Find the electric ±eld in the thick wire.
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Ch19-h1-solutions - pan (zp695) Ch19-h1 chiu (56565) This...

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