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Unformatted text preview: maldonado (om2892) – Homework 4 2010 UPII – Stone – (13750) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two identical parallel sections of metal rods are connected parallel to a battery as shown. The two sections of metal rods are free to move. b b 5 . 1m . 24 m 2 . 8 V Note: Neglect the resistance of the part of the circuit not specifically mentioned below. The battery is 2 . 8 V, each metal rod has a resistance of 2 . 04 Ω, a length of 5 . 1 m, and a mass of 0 . 0024 kg . The rods are separated by . 24 m . Find the magnitude of the acceleration of each rod. Correct answer: 0 . 00333606 m / s 2 . Explanation: Let : E = 2 . 8 V , R = 2 . 04 Ω , ℓ = 5 . 1 m , m = 0 . 0024 kg , and Δ r = 0 . 24 m . b b ℓ Δ r E The attractive forces on each metal rod are equal and opposite to each other, and each force has its magnitude given by F = I ℓ B with I the current in a rod, ℓ the length of that rod and B the magnetic field at the rod due to the current in the other rod. Since the separation Δ r of the rods, is much less than ℓ we can use Ampere’s law to find contintegraldisplay vector B · d vector ℓ = 2 π Δ r B = μ I B = μ I 2 π Δ r . Further, the voltage drop through the rod equals the voltage E of the battery; i.e. , E = I R , and I = E R , and B = μ parenleftbigg E R parenrightbigg 2 π Δ r . Substituting I = E R , and the expression for B , gives for the force, F = I ℓ B = parenleftbigg E R parenrightbigg 2 μ ℓ 2 π Δ r With F = ma , we find a = parenleftbigg E R parenrightbigg 2 μ ℓ 2 π Δ r m = parenleftbigg 2 . 8 V 2 . 04 Ω parenrightbigg 2 (1 . 25664 × 10 − 6 N / A 2 )(5 . 1 m) 2 π (0 . 24 m)(0 . 0024 kg) = . 00333606 m / s 2 . 002 (part 1 of 3) 10.0 points A circular current loop of radius R is placed in a horizontal plane and maintains a current I . There is a constant magnetic field vector B in the xy-plane, with the angle α ( α < 90 ◦ ) defined with respect to y-axis. The current in the loop flows counterclockwise as seen from above. maldonado (om2892) – Homework 4 2010 UPII – Stone – (13750) 2 I z ˆ k x ˆ ı y , ˆ B α What is the direction of the magnetic mo- ment vectorμ ? 1. +ˆ ı 2.- sin α ˆ - cos α ˆ ı 3. sin α ˆ + cos α ˆ ı 4. +ˆ correct 5.- cos α ˆ + sin α ˆ ı 6. cos α ˆ - sin α ˆ ı 7.- ˆ k 8. + ˆ k 9.- ˆ ı 10.- ˆ Explanation: I z ˆ k τ x ˆ ı μ loop B α According to the right-hand rule, the mag- netic moment is upward. 003 (part 2 of 3) 10.0 points If the current in the loop is 0 . 235 A and its radius is 1 . 67 cm, what is the magnitude of the magnetic moment of the loop?...
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This note was uploaded on 03/31/2011 for the course PHYS 1001 taught by Professor Stone during the Summer '10 term at Texas Brownsville.
- Summer '10