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Unformatted text preview: terry (ect328) – homework 28 – Turner – (59130) 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 yaxis and zaxis 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: 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 parenleftBig + ˆ k − ˆ ı parenrightBig 2. hatwide F = 1 √ 2 ( − ˆ ı +ˆ ) 3. hatwide F = 1 √ 2 (+ˆ ı − ˆ ) 4. vector F = 0 ; no deflection 5. hatwide F = 1 √ 2 parenleftBig − ˆ k +ˆ ı parenrightBig 6. hatwide F = 1 √ 2 ( − ˆ ı − ˆ ) 7. hatwide F = 1 √ 2 parenleftBig − ˆ k − ˆ ı parenrightBig 8. hatwide F = 1 √ 2 parenleftBig − ˆ k +ˆ parenrightBig 9. hatwide F = 1 √ 2 parenleftBig + ˆ k − ˆ parenrightBig correct 10. hatwide F = 1 √ 2 parenleftBig − ˆ k − ˆ 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 ı ) , vectorv = 1 √ 2 v parenleftBig +ˆ + ˆ k parenrightBig , and q < , therefore , vector F = − q  vectorv × vector B = − q  1 √ 2 v B bracketleftBigparenleftBig +ˆ + ˆ k parenrightBig × (+ˆ ı ) bracketrightBig = − q  1 √ 2 v B parenleftBig + ˆ k − ˆ parenrightBig hatwide F = 1 √ 2 parenleftBig + ˆ k − ˆ 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 : ℓ = 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 ℓ = m g B = m g I ℓ = (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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 Fall '09
 Turner
 Magnetic Field, Electric charge, Terry

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