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exam2sol

# exam2sol - Flores Michael Exam 2 Due Mar 9 2004 11:00 pm...

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Flores, Michael – Exam 2 – Due: Mar 9 2004, 11:00 pm – Inst: Sonia Paban 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points A resistor is made from a hollow cylinder of length l , inner radius a , and outer radius b . The region a < r < b is filled with material of resistivity ρ . Current runs along the axis of the cylinder. The resistance R of this component is 1. R = ρ ‘ π b 2 2. R = π b 2 ρ 3. R = π ( b 2 - a 2 ) ρ 4. R = ρ a π‘ b 2 5. R = ρ ‘ π a 2 6. R = ρ ‘ b 2 π a 4 7. R = 2 πρ ‘ ( b 2 - a 2 ) 8. R = ρ ‘ π ( b 2 - a 2 ) correct 9. R = ρ ‘ a 2 π b 4 10. R = 2 ρ ‘ π b 2 Explanation: By definition R = ρ A = ρ π b 2 - π a 2 = ρ ‘ π ( b 2 - a 2 ) . 002 (part 1 of 1) 10 points A particle with charge q and mass m is un- dergoing circular motion with speed v . At t = 0, the particle is moving along the nega- tive x axis in the plane perpendicular to the magnetic field ~ B , which points in the positive z direction in the figure below. x y z ~v ~ B Find the period T of oscillation; i.e. , the time it takes for the particle to complete one revolution. 1. T = m B q 2. T = 2 π m B q 3. T = q B m 4. T = m q B 5. T = 2 π m q B correct 6. T = 2 π q B m Explanation: Basic Concepts: Newton’s 2nd Law: F = ma Centripetal acceleration: F c = m v 2 r Force on charge q in magnetic field (no electric field): ~ F B = q~v × ~ B . The centripetal force F c is provided by F B , so m v 2 r = q v B , or v r = q B m .

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Flores, Michael – Exam 2 – Due: Mar 9 2004, 11:00 pm – Inst: Sonia Paban 2 the period of oscillation is T = 2 π r v , which is intuitive since the particle traverses a distance 2 π r in a revolution and, moving at speed v , takes the time T to do so. We know the ratio v r , so T = 2 π r v = 2 π m q B . 003 (part 1 of 1) 10 points A square loop of wire carries a current and is located in a uniform magnetic field. The left side of the loop is aligned and attached to a fixed axis (dashed line in figure). axis of rotation ~ B = 0 . 23 T ~ B = 0 . 23 T 0 . 44 m 4 . 6 A 0 . 44 m x y When the plane of the loop is parallel to the magnetic field in the position shown, what is the magnitude of the torque exerted on the loop about the axis of rotation, which is along the left side of the square as indicated by the dashed line in the figure? Correct answer: 0 . 204829 N m. Explanation: Let : d = 0 . 44 m , = 0 . 44 m , B = 0 . 23 T , and I = 4 . 6 A . axis of rotation ~ B ~ B d I x y Only the right side of the loop contributes to the torque. By definition, the torque is τ = k ~ d × ~ F k = d [ I ‘ B ] = (0 . 44 m) [(4 . 6 A) (0 . 44 m) (0 . 23 T)] = 0 . 204829 N m . 004 (part 1 of 1) 10 points A device (“source”) emits a bunch of charged ions (particles) with a range of velocities (see figure). Some of these ions pass through the left slit and enter region 1 in which there is a uniform electric field (vertically down) and a uniform magnetic field (into the page) as shown in the figure.
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