Homework 8-solutions

Homework 8-solutions - soto(rrs946 Homework 8 Homann(56605...

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soto (rrs946) – Homework 8 – Hoffmann – (56605) 1 This print-out should have 10 questions. Multiple-choice 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 x -axis 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

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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. x z A G a a Top View B B B ˆ ı ˆ k ˆ ı ˆ k B B B The magnetic force on a wire is given by vector F mag = I vector × vector B . The vector vector is given by the sum of the current segments vector = −→ AC + −→ CD + −→ DE + −→ EG , and this is the vector −→ AG , (see figure above).
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