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Unformatted text preview: Shie, Gary – Homework 19 – Due: Oct 18 2004, 4:00 am – Inst: Turner 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two long straight wires enter a room through a window 1.5 m high and 1.0 m wide. One carries a current of 7 . 7 A into the room while the other carries a current of 3 . 21 A out. Find the magnitude of I ~ B · ~ dl around the window frame. Correct answer: 5 . 6423 × 10 6 Tm. Explanation: According to Ampere’s Law, I ~ B · ~ dl = μ I , where I is the total current passing through the surface bounded by the closed path (here the window frame), I = (7 . 7 A) (3 . 21 A) = 4 . 49 A , so I ~ B · ~ dl = μ I = (4 π × 10 7 N / A 2 )(4 . 49 A) = 5 . 6423 × 10 6 Tm . 002 (part 1 of 1) 10 points A long, straight wire lies on a horizontal ta ble and carries a current of 2 . 48 μ A. In a vacuum, a proton moves parallel to the wire (opposite the current) with a constant veloc ity of 47200 m / s at a constant distance d above the wire. The acceleration of gravity is 9 . 8 m / s 2 . Find the value of d . You may ignore the magnetic field due to the Earth. Correct answer: 22 . 8876 cm. Explanation: Given : m = 1 . 67 × 10 27 kg , q = 1 . 6 × 10 19 C , v = 47200 m / s , and I = 2 . 48 μ A = 2 . 48 × 10 6 A . Since the proton moves with constant ve locity, the net force acting on it is zero. Thus, the magnetic force due to the current in the wire counterbalances the weight of the proton: q v B = mg B = μ I 2 π d q v μ I 2 π d = mg , so d = q v μ I 2 π mg = ( 1 . 6 × 10 19 C ) (47200 m / s) 2 π (1 . 67 × 10 27 kg)(9 . 8 m / s 2 ) × ( 1 . 25664 × 10 6 T · m / A ) × ( 2 . 48 × 10 6 A ) × 100 cm 1 m = 22 . 8876 cm . 003 (part 1 of 1) 10 points An infinitely long straight wire carrying a current I 1 = 28 . 2 A is partially surrounded by a loop as in figure. The loop has a length L = 50 . 1 cm, a radius R = 17 . 4 cm, and carries a current I 2 = 21 . 5 A. The axis of the loop coincides with the wire. R L I 1 I 2 Calculate the force exerted on the loop. Correct answer: 698 . 29 μ N. Explanation: Shie, Gary – Homework 19 – Due: Oct 18 2004, 4:00 am – Inst: Turner 2 The central wire creates field ~ B = μ I 1 2 π R counterclockwise . The curved portions of the loop feels zero force since ~ l × ~ B = 0 there. The straight portions both feel I 2 ~ l × ~ B forces to the right, amounting to ~ F = I 2 2 L μ I 1 2 π R = μ I 1 I 2 L π R to the right k ~ F k = μ (28 . 2 A)(21 . 5 A)(0 . 501 m) π (17 . 4 cm) = 698 . 29 μ N . 004 (part 1 of 1) 10 points Given: μ = 1 . 25664 × 10 6 N / A 2 ....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas.
 Spring '10
 ERSKINE
 Physics, Work

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