This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: nguyen (jmn727) homework 28 Turner (59070) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A positively charged particle moving at 45 angles to both the xaxis and yaxis enters a magnetic field (pointing into of the page), as shown in the figure below. x y v z vector B vector B + q Figure: is in the xdirection, is in the ydirection, and k is in the zdirection. What is the initial direction of deflection? 1. hatwide F = 1 2 parenleftBig + k parenrightBig 2. hatwide F = 1 2 parenleftBig + k parenrightBig 3. hatwide F = 1 2 parenleftBig + k + parenrightBig 4. hatwide F = 1 2 ( + ) correct 5. hatwide F = 1 2 parenleftBig k + parenrightBig 6. hatwide F = 1 2 parenleftBig + k parenrightBig 7. hatwide F = 1 2 (+ ) 8. hatwide F = 1 2 (+ + ) 9. vector F = 0 ; no deflection 10. hatwide F = 1 2 parenleftBig + + k parenrightBig Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv vector B Righthand rule for crossproducts. 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 parenleftBig + k parenrightBig , vectorv = 1 2 v (+ + ) , and q &gt; , therefore , vector F = +  q  vectorv vector B = +  q  1 2 v B bracketleftBig (+ + ) parenleftBig + k parenrightBigbracketrightBig = +  q  1 2 v B ( + ) hatwide F = 1 2 ( + ) . This is the second of eight versions of the problem. 002 10.0 points A thin 2.50 m long copper rod in a uniform magnetic field has a mass of 58.1 g. When the rod carries a current of 0.287 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 . 794371 T. Explanation: Let : = 2 . 50 m , m = 58 . 1 g = 0 . 0581 kg , I = 0 . 287 A , and g = 9 . 81 m / s 2 . The magnetic and gravitational forces are equal: F m = F g nguyen (jmn727) homework 28 Turner (59070) 2 B I = m g B = m g I = (0 . 0581 kg) (9 . 81 m / s 2 ) (0 . 287 A) (2 . 5 m) = . 794371 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....
View
Full Document
 Spring '08
 Turner
 Charge, Force, Work, Magnetic Field, Electric charge

Click to edit the document details