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# homework 5 - bhakta(nnb232 HW#5 Antoniewicz(56490 This...

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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 0 . 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. θ I B What is the force on the wire? Correct answer: 0 . 866025 N. Explanation: Let : I = 20 A , = 0 . 1 m , θ = 60 , and B = 0 . 5 T . we use F = I ℓ B sin θ , so F = I ℓ B sin θ = (20 A) (0 . 1 m) (0 . 5 T) sin 60 = 0 . 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 particle’s 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 Newton’s second law, F = q v B = m v 2 r v = q B r m . The kinetic energy is K = 1 2 m v 2 = q 2 B 2 r 2 2 m = qV , so the particle’s radius of curvature is r = 1 B radicalBigg 2 V m q = 1 0 . 073 T radicalBigg 2 (468 V) (6 . 6 × 10 27 kg) 3 . 20435 × 10 19 C = 0 . 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 ro n c o r e, p e r m ea b il i t y o f 4 0 0 0 μ 0 N t u r n s o f w i re i n t o r o id i

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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 = μ 0 i 2 π a 2. B = μ 0 i 2 a 3. B = μ 0 i 2 r 4. B = μ 0 i 2 b 5. B = 0 6. B = μ 0 N i 2 π b 7. B = μ 0 i 2 π r 8. B = μ 0 N i 2 π r correct 9. B = μ 0 N i 2 π a 10. B = μ 0 i 2 π 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 Ampere’s law gives 2 π r B = μ 0 N i . So B = μ 0 N i 2 π 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. The thumb gives the direction of the magnetic field inside of the toroid. When looking down from the top, one sees that the vector B -field lines form clockwise circular loops. 005 (part 3 of 3) 10.0 points Find the magnitude of the magnetic field in- side the central hole of the toroid at some point P = 1 2 h , where the perpendicular dis- tance from the central axis to the point P is 1 2 r .
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