homework 5 - bhakta (nnb232) HW #5 Antoniewicz (56490) 1...

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Unformatted text preview: bhakta (nnb232) HW #5 Antoniewicz (56490) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A wire carrying a current 20 A has a length . 1 m between the pole faces of a magnet at an angle 60 (see the figure). The magnetic field is approximately uniform at 0 . 5 T. We ignore the field beyond the pole pieces. I B What is the force on the wire? Correct answer: 0 . 866025 N. Explanation: Let : I = 20 A , = 0 . 1 m , = 60 , and B = 0 . 5 T . we use F = I B sin , so F = I B sin = (20 A) (0 . 1 m) (0 . 5 T) sin60 = . 866025 N . 002 10.0 points An alpha particle has a mass of 6 . 6 10 27 kg and is accelerated by a voltage of 0 . 468 kV. The charge on a proton is 1 . 60218 10 19 C. If a uniform magnetic field of 0 . 073 T is maintained on the alpha particle and perpen- dicular to its velocity, what will be particles radius of curvature? Correct answer: 0 . 0601474 m. Explanation: Let : B = 0 . 073 T , V = 0 . 468 kV = 468 V , m = 6 . 6 10 27 kg , and q = 2 e = 3 . 20435 10 19 C . From Newtons second law, F = q v B = mv 2 r v = q B r m . The kinetic energy is K = 1 2 mv 2 = q 2 B 2 r 2 2 m = qV , so the particles radius of curvature is r = 1 B radicalBigg 2 V m q = 1 . 073 T radicalBigg 2 (468 V) (6 . 6 10 27 kg) 3 . 20435 10 19 C = . 0601474 m . 003 (part 1 of 3) 10.0 points The toroid has its inner radius a , its outer radius b , a height of h , and its number of turns N . The rectangular cross-sectonal area of the hollow core is ( b a ) h . A r 2 r I r o n c o r e , p e rm ea bil it y o f 4 N t u r n s o f wi re in t o r o i d i bhakta (nnb232) HW #5 Antoniewicz (56490) 2 Find the magnitude of the magnetic field within a toroid at some point P = 1 2 h , where the perpendicular distance from the central axis to the point P is r . 1. B = i 2 a 2. B = i 2 a 3. B = i 2 r 4. B = i 2 b 5. B = 0 6. B = N i 2 b 7. B = i 2 r 8. B = N i 2 r correct 9. B = N i 2 a 10. B = i 2 b Explanation: Basic Concepts: Magnetic Field in Toroid. Denote the magnetic field at P by vector B , and construct a circular Amperian loop which is centered at the symmetry axis and it passes through P. Applying Amperes law gives 2 r B = N i . So B = N i 2 r . vector B is independent of the vertical height, as long as the point P is within the toroid. 004 (part 2 of 3) 10.0 points The direction of vector B field within the toroid as view from the top is 1. Cannot be determined. 2. counterclockwise 3. clockwise correct Explanation: Apply the right-hand rule. Curl the right- hand fingers along the direction of the current....
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This note was uploaded on 02/01/2011 for the course PHYSICS 302L taught by Professor Antanioweitz during the Fall '10 term at University of Texas at Austin.

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homework 5 - bhakta (nnb232) HW #5 Antoniewicz (56490) 1...

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