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Unformatted text preview: tovar (jdt436) – 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 negatively charged particle moving at 45 ◦ angles to both the zaxis and xaxis enters a magnetic field (pointing out of of the page), as shown in the figure below. z x v × y vector B vector B − q Figure: ˆ ı is in the xdirection, ˆ is in the ydirection, and ˆ k is in the zdirection. What is the initial direction of deflection? 1. hatwide F = 1 √ 2 ( − ˆ ı − ˆ ) 2. hatwide F = 1 √ 2 parenleftBig +ˆ + ˆ k parenrightBig 3. hatwide F = 1 √ 2 (+ˆ ı +ˆ ) 4. vector F = 0 ; no deflection 5. hatwide F = 1 √ 2 parenleftBig − ˆ ı + ˆ k parenrightBig 6. hatwide F = 1 √ 2 parenleftBig − ˆ ı − ˆ k parenrightBig 7. hatwide F = 1 √ 2 parenleftBig +ˆ ı + ˆ k parenrightBig correct 8. hatwide F = 1 √ 2 parenleftBig − ˆ − ˆ k parenrightBig 9. hatwide F = 1 √ 2 ( − ˆ ı +ˆ ) 10. hatwide F = 1 √ 2 parenleftBig − ˆ + ˆ 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 ( − ˆ ) , vectorv = 1 √ 2 v parenleftBig − ˆ k +ˆ ı parenrightBig , and q < , therefore , vector F = − q  vectorv × vector B = − q  1 √ 2 v B bracketleftBigparenleftBig − ˆ k +ˆ ı parenrightBig × ( − ˆ ) bracketrightBig = − q  1 √ 2 v B parenleftBig +ˆ ı + ˆ k parenrightBig hatwide F = 1 √ 2 parenleftBig +ˆ ı + ˆ k parenrightBig . This is the seventh of eight versions of the problem. 002 10.0 points A thin 2.66 m long copper rod in a uniform magnetic field has a mass of 57.4 g. When the rod carries a current of 0.264 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 . 801854 T. Explanation: Let : ℓ = 2 . 66 m , m = 57 . 4 g = 0 . 0574 kg , I = 0 . 264 A , and g = 9 . 81 m / s 2 . The magnetic and gravitational forces are equal: F m = F g tovar (jdt436) – homework 28 – Turner – (59070) 2 B I ℓ = m g B = m g I ℓ = (0 . 0574 kg) (9 . 81 m / s 2 ) (0 . 264 A) (2 . 66 m) = . 801854 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 12/11/2009 for the course PHY 303 taught by Professor Erskine/tsoi during the Spring '08 term at University of Texas at Austin.
 Spring '08
 ERSKINE/TSOI
 Charge, Force, Work

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