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# solution_pdf28 - tovar(jdt436 homework 28 Turner(59070 This...

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tovar (jdt436) – homework 28 – Turner – (59070) 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 z -axis and x -axis 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 x -direction, ˆ is in the y -direction, and ˆ k is in the z -direction. What is the initial direction of deflection? 1. hatwide F = 1 2 ( ˆ ı ˆ ) 2. hatwide F = 1 2 parenleftBig + 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 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 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 ( ˆ ) , vectorv = 1 2 v parenleftBig ˆ k ı parenrightBig , and q < 0 , therefore , vector F = −| q | vectorv × vector B = −| q | 1 2 v B bracketleftBigparenleftBig ˆ k ı parenrightBig × ( ˆ ) bracketrightBig = −| q | 1 2 v B parenleftBig ı + ˆ k parenrightBig hatwide F = 1 2 parenleftBig 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 : = 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

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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) = 0 . 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. a b c What is the direction of the magnetic field created by these wires at location a, b and c? (b is midway between the wires.) 1. up, down, up correct 2. up, up, down 3. up, zero, down 4. down, zero, up 5. down, up, down 6. down, down, up Explanation: By the right-hand rule the left wire has a clockwise field and the right wire a counter- clockwise field.
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