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# hw5 oscar - maldonado(om2892 Homework 4 2010 UPII...

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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. 5 . 1 m 0 . 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 0 . 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 , = 5 . 1 m , m = 0 . 0024 kg , and Δ r = 0 . 24 m . 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, 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 we can use Ampere’s law to find contintegraldisplay vector B · d vector = 2 π Δ r B = μ 0 I B = μ 0 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 0 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 0 2 π Δ r With F = m a , we find a = parenleftbigg E R parenrightbigg 2 μ 0 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) = 0 . 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.

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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. ı 2. - sin α ˆ - cos α ˆ ı 3. sin α ˆ + cos α ˆ ı 4. 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? Correct answer: 0 . 000205897 A · m 2 . Explanation: Let : I = 0 . 235 A , and R = 1 . 67 cm . The magnetic dipole moment is μ = I A = I π R 2 = (0 . 235 A) (3 . 15) (1 . 67 cm) 2 = 0 . 000205897 A · m 2 004 (part 3 of 3) 10.0 points What is the direction of the torque vector vector τ ? 1. - ˆ 2. - sin α ˆ - cos α ˆ ı 3. - cos α ˆ + sin α ˆ ı 4. sin α ˆ + cos α ˆ ı 5. ı 6. - ˆ ı 7. - ˆ k 8. cos α ˆ - sin α ˆ ı 9. + ˆ k correct 10. Explanation: The torque is vector τ = vectorμ × vector B = μ ) × [ B x ( - ˆ ı ) + B y (+ˆ )] and ˆ × ˆ = 0 and +ˆ × ( - ˆ ı ) =
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