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# 19 - Shie Gary Homework 19 Due 4:00 am Inst Turner This...

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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 T m. Explanation: According to Ampere’s Law, I ~ B · ~ dl = μ 0 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 0 I = (4 π × 10 - 7 N / A 2 ) (4 . 49 A) = 5 . 6423 × 10 - 6 T m . 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 = m g B = μ 0 I 2 π d q v μ 0 I 2 π d = mg , so d = q v μ 0 I 2 π m g = ( 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:

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Shie, Gary – Homework 19 – Due: Oct 18 2004, 4:00 am – Inst: Turner 2 The central wire creates field ~ B = μ 0 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 0 I 1 2 π R = μ 0 I 1 I 2 L π R to the right k ~ F k = μ 0 (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 0 = 1 . 25664 × 10 - 6 N / A 2 . A toroidal solenoid has an average radius of r = 10 . 6 cm and a cross sectional area of A = 2 . 49 cm 2 . There are N = 485 turns of wire on an iron core which has magnetic permeability of k μ 0 , where k = 4910. I A r Calculate the current necessary to produce a magnetic flux of Φ = 0 . 000798 Wb through a cross section of the core. Correct answer: 0 . 713274 A. Explanation: Basic concepts: Magnetic field strength H = N I 2 π r .
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