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# HW 23 - terry(ect328 homework 23 Turner(59130 This...

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terry (ect328) – homework 23 – Turner – (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An air-core solenoid has a length of 91 . 6 cm, a cross-sectional area of 0 . 0614 m 2 , and contains 152 turns. The current through the solenoid increases by 28 . 63 A. The permeability of free space is 1 . 25664 × 10 - 6 N / A 2 . By how much does the magnetic flux through the solenoid change? Correct answer: 366 . 562 μ Wb. Explanation: Let : = 91 . 6 cm = 0 . 916 m , A = 0 . 0614 m 2 , N = 152 turns , I = 28 . 63 A , and μ 0 = 1 . 25664 × 10 - 6 N / A 2 . The magnetic field inside the solenoid is B = μ 0 n I = (1 . 25664 × 10 - 6 N / A 2 ) × parenleftbigg 152 turns 0 . 916 m parenrightbigg (28 . 63 A) = 0 . 00597007 T . ΔΦ B = B A = (0 . 00597007 T) (0 . 0614 m 2 ) = 0 . 000366562 Wb = 366 . 562 μ Wb . 002 (part 1 of 2) 10.0 points Two separate loops each with the same radius and a current, in the directions shown, whose centers lie on the same axis perpendicular to the plane of the loops. 13 cm radius 455 A 455 A 8 . 8 m This figure is not to scale. The loop radii are small com- pared to the distance between the loops. Note: The left-hand side of the circular loops are closer to you. What approximately is the magnetic flux enclosed by the left loop due to the current in the right loop. Correct answer: 3 . 76415 × 10 - 10 Wb. Explanation: a radius I I Let : a = 13 cm , = 8 . 8 m , and I = 455 A . The magnitude of the magnetic field at the axis of a loop carrying a current I , if is much larger than a is given by B = μ 0 2 π μ 3 = μ 0 I a 2 2 3 = 2 π bracketleftBig μ 0 4 π bracketrightBig I a 2 3 , where the magnetic dipole moment μ = I A = I π a 2 . The magnetic field at the left hand loop due to the right hand loop is nearly constant and parallel to the axis. Thus the magnetic flux from this field is

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terry (ect328) – homework 23 – Turner – (59130) 2 Φ B = B A = B π a 2 = 2 π 2 bracketleftBig μ 0 4 π bracketrightBig I a 4 3 = 2 π 2 bracketleftbigg 4 π 10 - 7 T m / A 4 π bracketrightbigg × (455 A) (0 . 13 m) 4 (8 .
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HW 23 - terry(ect328 homework 23 Turner(59130 This...

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