PHY303LHW9S - rosen(arr956 – Homework 09 – Chiu...

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Unformatted text preview: rosen (arr956) – 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 2 . 06 cm and total charge 4 . 71 μ C rotates with a constant angular speed of 6 . 02 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: 6 . 01619 × 10- 9 A m 2 . Explanation: The period is T = 2 π ω . The spinning pro- duces a current I = Qf = Q T = Qω 2 π . Then, the magnetic moment μ is given by μ = I A = Qω 2 π π R 2 . Hence μ = QR 2 ω 2 = (4 . 71 × 10- 6 C) (0 . 0206 m) 2 (6 . 02 rad / s) 2 = 6 . 01619 × 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 . 009%. What is the magnetic susceptibility of the liquid? Correct answer: − 9 × 10- 5 . Explanation: Let : P = 0 . 009% = 9 × 10- 5 . The field inside the solenoid with the liquid sample present is B = B app (1 + χ m ) , where B app is the magnetic field in the ab- sence of the liquid sample. Thus the magnetic susceptibility χ m is χ m = Δ B B app = 9 × 10- 5 . 003 10.0 points A toroidal solenoid has an average radius of 14 . 26 cm and a cross sectional area of 2 . 689 cm 2 . There are 829 turns of wire on an iron core which has magnetic permeability of 4540 μ . The permeability of free space is 1 . 25664 × 10- 6 N / A 2 . I r cross sectional area is 2 . 689 cm 2 I r o n c o r e , p e rm ea bil it y o f 4 5 4 μ 8 2 9 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 . 000463 Wb through a cross section of the core. Correct answer: 0 . 326189 A. Explanation: Let : r = 14 . 26 cm = 0 . 1426 m , A = 2 . 689 cm 2 = 0 . 0002689 m 2 , N = 829 turns , μ = 1 . 25664 × 10- 6 N / A 2 , μ m = 4540 μ , and Φ m = 0 . 000463 Wb . Basic concepts: Magnetic field strength H = N I 2 π r . Magnetic flux density B = μ m H . Magnetic permeability μ m = k μ , rosen (arr956) – Homework 09 – Chiu – (58295) 2 where k = 1 + χ , and χ is the susceptibility. Solution: To determine the magnetic field strength H in free space H = N I 2 π r , where r is radius of toroid. Thus magnetic flux density B is B = μ m H = μ m N I 2 π r . Assuming no vector B field outside the solenoid and uniform field inside, the magnetic flux is Φ m = B A = μ m N I A 2 π r Therefore, the required current is given by I = Φ m 2 π r μ m N A = Φ m 2 π r k μ N A = (0 . 000463 Wb) 2 π (4540)(1 . 25664 × 10- 6 N / A 2 ) × (0 . 1426 m) (829 turns)(0 . 0002689 m 2 ) = . 326189 A ....
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PHY303LHW9S - rosen(arr956 – Homework 09 – Chiu...

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