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Unformatted text preview: yang (ey942) – ohw22 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points The magnetic field is constant and it points into the paper. The instantaneous velocity vector of a positive charge particle is pointing to the right. + + + + + + + + + + + + + + + + q v The trajectory of the subsequent motion is 1. bending along a counterclockwise path. correct 2. bending along a clockwise path. 3. to continue to move along a straight line to the right. Explanation: vector F = q vector v × vector B vector F = I vector l × vector B E =- d Φ dt The particle experiences Lorenz force vector F = q vector v × vector B , which is pointing upward at that instant. So the trajectory of subsequent motion is coun- terclockwise. 002 (part 2 of 4) 10.0 points Two parallel wires carry opposite current as shown. i 1 i 2 Find the direction of the magnetic force on i 2 due to the magnetic field of i 1 . 1. into the paper 2. to the right correct 3. out of the paper 4. to the left Explanation: i 1 i 2 F 2 B 1 Wire 1, which carries a current i 1 , sets up a magnetic field vector B 1 , which points into the paper at the position of wire 2. The direction of vector B 1 is perpendicular to the wire, as illustrated below. The magnetic force on a length l of wire 2 is vector F 2 = i 2 vector l × vector B 1 . Since i 2 flows downward, vector l × vector B 1 is to the right. 003 (part 3 of 4) 10.0 points A metal bar pivoting at A is rotating in a uniform magnetic field (see the setup shown below). yang (ey942) – ohw22 – turner – (56705) 2 A B v The potential difference V A- V B is 1. positive. correct 2. negative. 3. zero. Explanation: v q A B Since the electrons move with the bar, the direction of their velocity is the same as the direction of rotation. vector F = q vector v × vector B Noting that the charge of electrons is nega- tive, we find out that the magnetic force is pointing radially outwards (see the figure be- low). Since electrons move radially outwards, nega- tive net charge will cumulate at the outward end of the metal bar, while there will be pos- itive net charge at the inward end. Therefore the potential V A should be higher than V B . Alternatively, the magnetic force on positive charge is directed radially inward. Thus the magnetic force does work on positive charge in pushing it from B to A. In analogy to ski lift, the magnetic force acts as the external force in lifting the “chair-lift” from lower poten- tial point to higher potential point. Therefore V A > V B , or V A- V B is positive. 004 (part 4 of 4) 10.0 points Imagine that Galileo had dropped a bar mag- net down the copper tube from the Tower of Pisa....
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This note was uploaded on 11/21/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.
- Spring '10