19 - Shie, Gary Homework 19 Due: Oct 18 2004, 4:00 am Inst:...

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Unformatted text preview: Shie, Gary Homework 19 Due: Oct 18 2004, 4:00 am Inst: Turner 1 This print-out should have 12 questions. Multiple-choice 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 Amperes Law, I ~ B ~ dl = 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 = I = (4 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 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 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 = I 2 d q v I 2 d = mg , so d = q v I 2 mg = ( 1 . 6 10- 19 C ) (47200 m / s) 2 (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 N. Explanation: Shie, Gary Homework 19 Due: Oct 18 2004, 4:00 am Inst: Turner 2 The central wire creates field ~ B = I 1 2 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 I 1 2 R = I 1 I 2 L R to the right k ~ F k = (28 . 2 A)(21 . 5 A)(0 . 501 m) (17 . 4 cm) = 698 . 29 N . 004 (part 1 of 1) 10 points Given: = 1 . 25664 10- 6 N / A 2 ....
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19 - Shie, Gary Homework 19 Due: Oct 18 2004, 4:00 am Inst:...

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