PHY303L Solutions%209

PHY303L Solutions%209 - homework 09 – FAUSAK TAYLOR –...

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Unformatted text preview: homework 09 – FAUSAK, TAYLOR – Due: Mar 24 2008, 4:00 am 1 Question 1, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the figure, the resistor is 5 Ω and a 8 T magnetic field is directed into the paper. The separation between the rails is 8 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 7 m / s . Assume the bar and rails have negligible resistance and friction. m ≪ 1g 7 m / s 5Ω 8 T 8 T I 8m Calculate the applied force required to move the bar to the left at a constant speed of 7 m / s. Correct answer: 5734 . 4 N (tolerance ± 1 %). Explanation: Motional emf is E = B ℓ v . Magnetic force on current is vector F = I vector ℓ × vector B . Ohm’s Law is I = V R . The motional emf induced in the circuit is E = B ℓ v = (8 T) (8 m) (7 m / s) = 448 V . From Ohm’s law, the current flowing through the resistor is I = E R = 448 V 5 Ω = 89 . 6 A . Thus, the magnitude of the force exerted on the bar due to the magnetic field is F B = I ℓ B = (89 . 6 A)(8 m)(8 T) = 5734 . 4 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = F B = 5734 . 4 N . Question 2, chap 31, sect 1. part 2 of 2 10 points At what rate is energy dissipated in the resistor? Correct answer: 40140 . 8 W (tolerance ± 1 %). Explanation: The power dissipated in the resistor is P = I 2 R = (89 . 6 A) 2 (5 Ω) = 40140 . 8 W . Note: First of four versions. Question 3, chap 31, sect 1. part 1 of 2 10 points The figure below shows one of the blades of a helicopter which rotates around a central hub. The vertical component of the Earth’s magnetic field is into the plane of the paper. 2 . 3 m 1 × 10 − 5 T 1 × 10 − 5 T 2 . 3 r e v / s homework 09 – FAUSAK, TAYLOR – Due: Mar 24 2008, 4:00 am 2 What is the magnitude of the emf E induced between the blade tip and the central hub? Correct answer: 0 . 000382238 V (tolerance ± 1 %). Explanation: ℓ B B f Let : ℓ = 2 . 3 m , f = 2 . 3 rev / s , and B = 1 × 10 − 5 T . For a point on the blade, the velocity with which the point moves changes linearly with the distance from the point to the center of the hub. Then the effective velocity for the whole blade is the mean velocity, v eff = ω · ℓ 2 = 2 πf · ℓ 2 = 2 π (2 . 3 rev / s)(2 . 3 m) 2 = 16 . 619 m / s , and the induced emf in the blade is E = B ℓ v eff = 1 2 B ω ℓ 2 = 1 2 (1 × 10 − 5 T) × (14 . 4513 rad / s) (2 . 3 m) 2 = 0 . 000382238 V . Alternative Solution: An alternative method is to calculate d Φ B dt = B d A dt , where A is the area enclosed in the diagram below and Φ B the enclosed flux in the figure. ℓ ω θ The area inside the triangle is A = ℓ 2 2 θ , and d A d t = ℓ 2 2 d θ dt = ω ℓ 2 2 ....
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PHY303L Solutions%209 - homework 09 – FAUSAK TAYLOR –...

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