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# HW 5 - segovia(cs39966 Homework 5 Spurlock(14441 This...

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segovia (cs39966) – Homework 5 – Spurlock – (14441) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A circular coil consisting of a single loop of wire has a radius of 19 . 6 cm and carries a current of 37 . 3 A. It is placed in an external magnetic field of 0 . 258 T. Find the magnitude of the torque on the wire when the plane of the coil makes an angle of 25 . 3 with the direction of the field. Correct answer: 1 . 05002 N · m. Explanation: Let : r = 19 . 6 cm = 0 . 196 m , B = 0 . 258 T , I = 37 . 3 A , and θ = 25 . 3 . The area of the circular coil is A = π r 2 = π (0 . 196 m) 2 = 0 . 120687 m 2 . The torque is bardbl vector τ bardbl = bardbl vectorμ × vector B bardbl = μ B sin θ , where θ is the angle that the normal to the loop makes with the magnetic field, and μ = I A is the magnetic moment of the loop. Thus the magnitude of the torque is τ = I A B sin θ = (37 . 3 A) (0 . 120687 m 2 ) (0 . 258 T) × sin(90 25 . 3 ) = 1 . 05002 N · m . 002 10.0points Two long straight wires carry currents per- pendicular to the xy plane. One wire carries a current of 97 A and passes through the point x = 11 cm on the x axis. The second wire car- ries a current of 12000 A and passes through the point y = 7 . 1 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin? Correct answer: 33803 . 3 μ T. Explanation: Let : x = 11 cm = 0 . 11 m , I 1 = 97 A , y = 7 . 1 cm = 0 . 071 m , I 2 = 12000 A , and μ 0 = 4 π × 10 7 T · m / A . The magnetic field from the first wire will be vector B 1 = μ 0 I 1 2 π x ( ± ˆ ) directed along the y axis at the origin. The magnetic field from the second wire will be vector B 2 = μ 0 I 2 2 π y ( ± ˆ ı ) directed along the x axis at the origin. Since these contributions are perpendicular to each other, the magnitude of the total magnetic field at the origin will be B = radicalBig B 2 1 + B 2 2 = radicalBigg parenleftbigg μ 0 I 1 2 π x parenrightbigg 2 + parenleftbigg μ 0 I 2 2 π y parenrightbigg 2 = radicalBigg parenleftBig μ 0 2 π parenrightBig 2 parenleftbigg I 2 1 x 2 + I 2 2 y 2 parenrightbigg = μ 0 2 π radicalBigg I 2 1 x 2 + I 2 2 y 2 = 4 π × 10 7 T · m / A 2 π × radicalBigg (97 A) 2 (0 . 11 m) 2 + (12000 A) 2 (0 . 071 m) 2 × 10 6 μ T 1 T = 33803 . 3 μ T .

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segovia (cs39966) – Homework 5 – Spurlock – (14441) 2 003 10.0points A long, thin conductor carries a current of 10 . 5 A. At what distance from the conductor is the magnitude of the resulting magnetic field 0 . 000114 T? Correct answer: 1 . 84211 cm. Explanation: Let : μ 0 = 1 . 25664 × 10 6 N / A 2 , B = 0 . 000114 T , and I = 10 . 5 A . The magnetic field is B = μ 0 I 2 π r r = μ 0 I 2 π B = (1 . 25664 × 10 6 N / A 2 ) (10 . 5 A) 2 π (0 . 000114 T) = 0 . 0184211 m = 1 . 84211 cm . keywords: 004 10.0points Two long, straight parallel wires 8 . 2 cm apart carry currents of equal magnitude I . The parallel wires repel each other with a force per unit length of 2 . 8 nN / m. The permeability of free space is 4 π × 10 7 T · m / A.
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HW 5 - segovia(cs39966 Homework 5 Spurlock(14441 This...

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