Homework_09-solutions

# Homework_09-solutions - toupal (rgt374) Homework 09 Chiu...

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Unformatted text preview: toupal (rgt374) Homework 09 Chiu (58295) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A uniform non-conducting ring of radius 1 . 71 cm and total charge 6 . 46 C rotates with a constant angular speed of 4 . 3 rad / s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring? Correct answer: 4 . 06128 10- 9 A m 2 . Explanation: The period is T = 2 . The spinning pro- duces a current I = Qf = Q T = Q 2 . Then, the magnetic moment is given by = I A = Q 2 R 2 . Hence = QR 2 2 = (6 . 46 10- 6 C) (0 . 0171 m) 2 (4 . 3 rad / s) 2 = 4 . 06128 10- 9 A m 2 . 002 10.0 points When a sample of liquid is inserted into a solenoid carrying a constant current, the mag- netic field inside the solenoid decreases by . 012%. What is the magnetic susceptibility of the liquid? Correct answer: . 00012. Explanation: Let : P = 0 . 012% = 0 . 00012 . The field inside the solenoid with the liquid sample present is B = B app (1 + m ) , where B app is the magnetic field in the ab- sence of the liquid sample. Thus the magnetic susceptibility m is m = B B app = . 00012 . 003 10.0 points A toroidal solenoid has an average radius of 10 . 44 cm and a cross sectional area of 2 . 141 cm 2 . There are 700 turns of wire on an iron core which has magnetic permeability of 4560 . The permeability of free space is 1 . 25664 10- 6 N / A 2 . I r cross sectional area is 2 . 141 cm 2 I r o n c o r e , p e rm ea bil it y o f 4 5 6 7 t u r n s t u r n s o f w ir e i n t o r o i d Calculate the current necessary to produce a magnetic flux of 0 . 00057 Wb through a cross section of the core. Correct answer: 0 . 435377 A. Explanation: Let : r = 10 . 44 cm = 0 . 1044 m , A = 2 . 141 cm 2 = 0 . 0002141 m 2 , N = 700 turns , = 1 . 25664 10- 6 N / A 2 , m = 4560 , and m = 0 . 00057 Wb . Basic concepts: Magnetic field strength H = N I 2 r . Magnetic flux density B = m H . Magnetic permeability m = k , where k = 1 + , and is the susceptibility. toupal (rgt374) Homework 09 Chiu (58295) 2 Solution: To determine the magnetic field strength H in free space H = N I 2 r , where r is radius of toroid. Thus magnetic flux density B is B = m H = m N I 2 r . Assuming no vector B field outside the solenoid and uniform field inside, the magnetic flux is m = B A = m N I A 2 r Therefore, the required current is given by I = m 2 r m N A = m 2 r k N A = (0 . 00057 Wb) 2 (4560)(1 . 25664 10- 6 N / A 2 ) (0 . 1044 m) (700 turns)(0 . 0002141 m 2 ) = . 435377 A ....
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## This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas at Austin.

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Homework_09-solutions - toupal (rgt374) Homework 09 Chiu...

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