oldhw18_pdf - mcconnell (kam2342) – oldhomework 18 –...

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Unformatted text preview: mcconnell (kam2342) – 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 200-turn circular coil of radius 0 . 6 cm that carries a current of 1 . 5 mA and resides in a uniform magnetic field of 0 . 4 T? Correct answer: 1 . 35717 × 10 − 5 N · m. Explanation: Let : N = 200 , B = 0 . 4 T , r = 0 . 6 cm = 0 . 006 m , and I = 1 . 5 mA = 0 . 0015 A . μ = N I A = N I π r 2 τ max = μ B = N I π r 2 B = (200) (0 . 0015 A) π × (0 . 006 m) 2 (0 . 4 T) = 1 . 35717 × 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 2. a = q v m B 3. a = q m v B 4. a = v B q m 5. a = q v B m correct 6. a = q B m v Explanation: Basic Concepts: Newton’s 2nd Law: F = ma . Centripetal acceleration: F c = mv 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 = ma and F B = q v B , the particle moves perpendicularly to the mag- netic field, so that ma = 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 = mB q mcconnell (kam2342) – oldhomework 18 – Turner – (60230) 2 4. T = m q B 5. T = 2 π q B m 6. T = 2 π mB q Explanation: The centripetal force F c is provided by F B , so mv 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 k 2. hatwide a = ˆ k 3. hatwide a = ˆ j 4. hatwide a = − ˆ j correct 5. hatwide a = ˆ i 6. hatwide a = − ˆ i 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....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.

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oldhw18_pdf - mcconnell (kam2342) – oldhomework 18 –...

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