exam4sol

# exam4sol - Flores, Michael – Exam 4 – Due: Apr 23 2004,...

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Unformatted text preview: Flores, Michael – Exam 4 – Due: Apr 23 2004, noon – Inst: Sonia Paban 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A metal bar with length OA = L is rotating in a counterclockwise manner about the point O with a constant angular velocity ω . There is a constant magnetic field, B , directed out of the paper. A positive charge q is fixed at a point C on the rod. L v A O C The direction of the magnetic force on the charge q at point C due to the magnetic field, and the relationship between V O and V A are respectively given by: (Caution: o is the pivot point of rotation.) 1. radially inward, V A > V O 2. out of the plane, V O = V A 3. radially inward, V O > V A 4. radially outward, V O > V A 5. radially outward, V A > V O correct 6. radially outward, V O = V A 7. in the direction of ~ v , V O = V A 8. in the plane, V O = V A 9. opposite to the direction of ~ v , V O = V A Explanation: v A O B F B From the figure above we can see that the force on q is directed radially outward. This may be obtained from the equation ~ F = q ~ v × ~ B Because of this magnetic force, the positive charges begin to accumulate at A (or nor- mally, the negative charges begin to accumu- late at O ), producing an electric field that points from A to O . Hence, V A > V O . 002 (part 1 of 1) 10 points An inductor in the form of an air-core solenoid contains 177 turns, is of length 13 . 2 cm, and has a cross-sectional area of 4 . 51 cm 2 ....
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## This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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exam4sol - Flores, Michael – Exam 4 – Due: Apr 23 2004,...

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