PHY303L Solutions%209 - homework 09 FAUSAK, TAYLOR Due: Mar...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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 5 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 v . Magnetic force on current is vector F = I vector vector B . Ohms Law is I = V R . The motional emf induced in the circuit is E = B v = (8 T) (8 m) (7 m / s) = 448 V . From Ohms law, the current flowing through the resistor is I = E R = 448 V 5 = 89 . 6 A . Thus, the magnitude of the force exerted on the bar due to the magnetic field is F B = I 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 ) = 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 Earths 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: B B f Let : = 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 = 2 = 2 f 2 = 2 (2 . 3 rev / s)(2 . 3 m) 2 = 16 . 619 m / s , and the induced emf in the blade is E = B v eff = 1 2 B 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 B dt = B d A dt , where A is the area enclosed in the diagram below and B the enclosed flux in the figure. The area inside the triangle is A = 2 2 , and d A d t = 2 2 d dt = 2 2 ....
View Full Document

Page1 / 11

PHY303L Solutions%209 - homework 09 FAUSAK, TAYLOR Due: Mar...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online