This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: soto (rrs946) – Homework 8 – Hoffmann – (56605) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle with charge q and mass m has speed v . At t = 0, the particle is moving along the negative x axis in the plane perpendicular to the magnetic field vector B , which points in the positive z direction in the figure below. x y z vectorv vector B Find the direction of the instantaneous ac celeration hatwide a at t = 0 if q is negative. 1. hatwide a = ˆ j 2. hatwide a = ˆ j + ˆ k 3. hatwide a = − ˆ k + ˆ i 4. hatwide a = − ˆ j correct 5. hatwide a = − ˆ i 6. hatwide a = ˆ k 7. hatwide a = ˆ i 8. hatwide a = − ˆ k 9. hatwide a = ˆ k + ˆ i 10. hatwide a = ˆ i + ˆ j Explanation: The particle is moving along the negative xaxis in this instant vectorv = − v ˆ i ; since it is moving in a circle, we need to talk about instantaneous direction. The force F B is equal to q vectorv × vector B at all times. We know that vector B is pointing in the z direction, so vector B = B ˆ k , and therefore vector F B = q v ( − ˆ i ) × B ˆ k = q v B ( − ˆ i × ˆ k ) = q v B ˆ j . The charge q is negative ( q = − q  ) , so vector F B = − q  v B ˆ j =  q  v B ( − ˆ j ) . All quantities are positive, so the actual di rection in which vector F B points is the negative y direction, or hatwide a = − ˆ j . 002 10.0 points Given a current segment which flows along the edges of a cube as shown in the figure. The conventional Cartesian notation of ˆ ı (a unit vector along the positive x axis), ˆ (a unit vector along the positive y axis), and ˆ k (a unit vector along the positive z axis), is used. The cube has sides of length a . The current flows along the path A → C → D → E → G . There is a uniform magnetic field vector B = B ˆ ı . x y z B B B A C D E G a a Find the direction hatwide F ≡ vector F bardbl vector F bardbl of the resul tant magnetic force on the current segment ACDEG . 1. hatwide F = ˆ 2. hatwide F = − ˆ correct 3. hatwide F = 1 √ 2 parenleftBig ˆ − ˆ k parenrightBig soto (rrs946) – Homework 8 – Hoffmann – (56605) 2 4. hatwide F = 1 √ 2 parenleftBig ˆ k − ˆ parenrightBig 5. hatwide F = ˆ ı 6. hatwide F = ˆ k 7. hatwide F = − ˆ k 8. hatwide F = 1 √ 2 parenleftBig ˆ + ˆ k parenrightBig 9. Undetermined, since the magnitude of the force is zero. 10. hatwide F = − ˆ ı Explanation: Note: The force on wire segment AC is canceled by the force on wire segment EG . The current in wire segment CD flows in the ˆ ı direction and the current in wire segment DE flows in the − ˆ k direction....
View
Full
Document
This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.
 Fall '11
 HOFFMAN
 Force, Work

Click to edit the document details