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Unformatted text preview: Niu (qn269) – Homework #5 – antoniewicz – (58855) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A wire carrying a current 40 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. ℓ θ I B What is the force on the wire? Correct answer: 1 . 73205 N. Explanation: Let : I = 40 A , ℓ = 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 = (40 A) (0 . 1 m) (0 . 5 T) sin60 ◦ = 1 . 73205 N . 002 (part 1 of 2) 6.0 points In the figure, a 40 cm length of conducting wire that is free to move is held in place between two thin conducting wires. All of the wires are in a magnetic field. When a 11 A current is in the wire, as shown in the figure, the wire segment moves upward at a constant velocity. The acceleration of gravity is 9 . 81 m / s 2 . 40 cm 11 A 11 A 11 A a) Assuming the wire slides without friction on the two vertical conductors and has a mass of 0 . 13 kg, find the magnitude of the minimum magnetic field that is required to move the wire. Correct answer: 0 . 289841 T. Explanation: Let : ℓ = 40 cm , I = 11 A , m = 0 . 13 kg , and g = 9 . 81 m / s 2 . The magnetic and gravitational forces are equal: F m = F g B I ℓ = mg B = mg I ℓ = (0 . 13 kg) (9 . 81 m / s 2 ) (11 A) (0 . 4 m) = . 289841 T 003 (part 2 of 2) 4.0 points b) What is its direction? 1. Into the page 2. None of these 3. Toward the left edge of the page 4. Toward the top edge of the page 5. Toward the bottom edge of the page 6. Out of the page correct 7. Toward the right edge of the page Niu (qn269) – Homework #5 – antoniewicz – (58855) 2 Explanation: Apply righthand rule; force directed out of the palm of the hand, fingers in the direction of the field, thumb in the direction of the current. Thumb points to the left, palm faces toward the top of the page, and the fingers point out of the page. 004 10.0 points An alpha particle has a mass of 6 . 6 × 10 − 27 kg and is accelerated by a voltage of 1 . 35 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 . 102155 m. Explanation: Let : B = 0 . 073 T , V = 1 . 35 kV = 1350 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 = 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 particle’s radius of curvature is r = 1 B radicalBigg 2 V m q = 1 . 073 T radicalBigg 2 (1350 V) (6 . 6 × 10 − 27 kg) 3 . 20435 × 10 − 19 C = . 102155 m ....
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This note was uploaded on 11/18/2009 for the course PHY 58855 taught by Professor Antoniewicz during the Spring '09 term at University of Texas.
 Spring '09
 Antoniewicz
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