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Unformatted text preview: toupal (rgt374) – Homework 08 – Chiu – (58295) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0 . 9 m long wire weighing 0 . 072 N / m is suspended directly above an infinitely straight wire. The top wire carries a current of 47 A and the bottom wire carries a current of 82 A . The permeablity of free space is 1 . 25664 × 10 − 6 N / A 2 . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Correct answer: 10 . 7056 mm. Explanation: Let : I 1 = 47 A , I 2 = 82 A , λ f = 0 . 072 N / m , and μ = 1 . 25664 × 10 − 6 N / A 2 . In order for the system to be in equilibrium, the magnetic force per unit length λ F on the top wire must be equal to its weight per unit length. Thus λ F = F L = μ I 1 I 2 2 π d so d = μ I 1 I 2 2 π λ F = (1 . 25664 × 10 − 6 N / A 2 ) (47 A) (82 A) 2 π (0 . 072 N / m) = 0 . 0107056 m = 10 . 7056 mm . 002 10.0 points A rectangular loop with dimensions (hor izontal = 0 . 11 m) × (vertical= 0 . 253 m), is suspended by a string, and the lower hori zontal section of the loop is immersed in a magnetic field. If a current of 6 A is maintained in the loop, what is the magnitude of the magnetic field required to produce a tension of 0 . 035 N in the supporting string? Assume: Gravitational force is negligible. Correct answer: 0 . 0530303 T. Explanation: Let : F = 0 . 035 N , I = 6 A , and l hor = 0 . 11 m . The magnitude of the force on the lower leg is  vector F  =  I vector ℓ hor × vector B  . Therefore, the magnitude of the magnetic field due to this force is B = F I ℓ hor = . 035 N (6 A) (0 . 11 m) = 0 . 0530303 T . 003 (part 1 of 2) 10.0 points Consider a current in the long, straight wire which lies in the plane of the rectangular loop, that also carries a current, as shown. toupal (rgt374) – Homework 08 – Chiu – (58295) 2 11 cm 13 cm 66cm 4 . 3A → 15A → x y Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Correct answer: 4 . 1925 × 10 − 5 N. Explanation: Let : c = 0 . 11 m , a = 0 . 13 m , ℓ = 0 . 66 m , I 1 = 4 . 3 A , and I 2 = 15 A . c a ℓ I 1 → I 2 → x y The magnetic forces on the top and the bottom segments of the rectangle cancel. We now concern ourselves with the seg ments of the rectangle which are parallel to the long wire. Using Ampere’s law, the mag netic fields from the long wire at distances c and ( c + a ) away are B c = − μ I 1 2 π c ˆ k B ca = − μ I 1 2 π ( c + a ) ˆ k . The forces on the left and right vertical segments of the rectangle are, respectively, F c = I 2 B c ℓ and F ca = I 2 B ca ℓ . The forces are oppositely directed since the current I 2 has a direction in the loop at distances c op posite from the current I 2 in the loop at a distance c + a , away from the long wire....
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 Spring '09
 Chiu
 Physics, Electron, Force, Work, Magnetic Field, Electric charge

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