18 - Kapoor(mk9499 oldhomework 18 Turner(60230 This...

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Kapoor (mk9499) – oldhomework 18 – Turner – (60230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the maximum torque on a 450-turn circular coil of radius 0 . 85 cm that carries a current of 1 mA and resides in a uniform magnetic field of 0 . 2 T? Correct answer: 2 . 04282 × 10 5 N · m. Explanation: Let : N = 450 , B = 0 . 2 T , r = 0 . 85 cm = 0 . 0085 m , and I = 1 mA = 0 . 001 A . μ = N I A = N I π r 2 τ max = μ B = N I π r 2 B = (450) (0 . 001 A) π × (0 . 0085 m) 2 (0 . 2 T) = 2 . 04282 × 10 5 N · m . 002 (part 1 of 3) 10.0 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 vector B , which points in the positive z direction (up from the plane of the figure). x y z vectorv vector B Find the magnitude of the centripetal ac- celeration ( bardbl vectora bardbl ≡ a ). 1. a = q v B m correct 2. a = q B m v 3. a = q m v B 4. a = v B q m 5. a = q v B m 6. a = q v m B Explanation: Basic Concepts: Newton’s 2nd Law: F = m a . Centripetal acceleration: F c = m v 2 r Force on charge q in magnetic field (no electric field): vector F B = qvectorv × vector B . Solution: The trajectory of the particle will be bent into a circle by the magnetic field. From this we understand there has to be a centripetal acceleration a to keep the particle in this circle (just like a string is needed to provide the tension to keep a ball whirling in a circle). Since F = m a and F B = q v B , the particle moves perpendicularly to the mag- netic field, so that m a = q v B , a = q v B m . 003 (part 2 of 3) 10.0 points Find the period T of oscillation; i.e. , the time it takes for the particle to complete one revo- lution. 1. T = q B m 2. T = 2 π m q B correct 3. T = 2 π m B q
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Kapoor (mk9499) – oldhomework 18 – Turner – (60230) 2 4. T = 2 π q B m 5. T = m q B 6. T = m B q Explanation: The centripetal force F c is provided by F B , so m v 2 r = q v B , v r = q B m . 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 . 004 (part 3 of 3) 10.0 points Find the direction of the instantaneous accel- eration hatwide a at t = 0 if q is negative. 1. hatwide a = ˆ j correct 2. hatwide a = ˆ k 3. hatwide a = ˆ i 4. hatwide a = ˆ k 5. hatwide a = ˆ i 6. hatwide a = 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 qvectorv × 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 .
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