HW 28 - terry (ect328) homework 28 Turner (59130) 1 This...

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Unformatted text preview: terry (ect328) homework 28 Turner (59130) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A negatively charged particle moving at 45 angles to both the y-axis and z-axis enters a magnetic field (pointing into of the page), as shown in the figure below. y z v x vector B vector B q Figure: is in the x-direction, is in the y-direction, and k is in the z-direction. What is the initial direction of deflection? 1. hatwide F = 1 2 parenleftBig + k parenrightBig 2. hatwide F = 1 2 ( + ) 3. hatwide F = 1 2 (+ ) 4. vector F = 0 ; no deflection 5. hatwide F = 1 2 parenleftBig k + parenrightBig 6. hatwide F = 1 2 ( ) 7. hatwide F = 1 2 parenleftBig k parenrightBig 8. hatwide F = 1 2 parenleftBig k + parenrightBig 9. hatwide F = 1 2 parenleftBig + k parenrightBig correct 10. hatwide F = 1 2 parenleftBig k parenrightBig Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv vector B Right-hand rule for cross-products. hatwide F vector F bardbl vector F bardbl ; i.e. , a unit vector in the F direc- tion. Solution: The force is vector F = qvectorv vector B . vector B = B (+ ) , vectorv = 1 2 v parenleftBig + + k parenrightBig , and q < , therefore , vector F = | q | vectorv vector B = | q | 1 2 v B bracketleftBigparenleftBig + + k parenrightBig (+ ) bracketrightBig = | q | 1 2 v B parenleftBig + k parenrightBig hatwide F = 1 2 parenleftBig + k parenrightBig . This is the fourth of eight versions of the problem. 002 10.0 points A thin 2.54 m long copper rod in a uniform magnetic field has a mass of 42.8 g. When the rod carries a current of 0.218 A, it floats in the magnetic field. The acceleration of gravity is 9 . 81 m / s 2 . What is the field strength of the magnetic field? Correct answer: 0 . 758268 T. Explanation: Let : = 2 . 54 m , m = 42 . 8 g = 0 . 0428 kg , I = 0 . 218 A , and g = 9 . 81 m / s 2 . The magnetic and gravitational forces are equal: F m = F g terry (ect328) homework 28 Turner (59130) 2 B I = m g B = m g I = (0 . 0428 kg) (9 . 81 m / s 2 ) (0 . 218 A) (2 . 54 m) = . 758268 T 003 10.0 points The figure represents two long, straight, par- allel wires extending in a direction perpendic- ular to the page. The current in the left wire runs into the page and the current in the right runs out of the page....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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HW 28 - terry (ect328) homework 28 Turner (59130) 1 This...

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