Homework_09-solutions

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

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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 – fnd all choices beFore answering. 001 10.0 points A uniForm non-conducting ring oF radius 1 . 71 cm and total charge 6 . 46 μ 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 Am 2 . Explanation: The period is T = 2 π ω . The spinning pro- duces a current I = Qf = Q T = 2 π . Then, the magnetic moment μ is given by μ = IA = 2 π π R 2 . Hence μ = QR 2 ω 2 = (6 . 46 × 10 - 6 C) (0 . 0171 m) 2 (4 . 3 rad / s) 2 = 4 . 06128 × 10 - 9 2 . 002 10.0 points When a sample oF liquid is inserted into a solenoid carrying a constant current, the mag- netic feld inside the solenoid decreases by 0 . 012%. What is the magnetic susceptibility oF the liquid? Correct answer: - 0 . 00012. Explanation: Let : P =0 . 012% = 0 . 00012 . The feld inside the solenoid with the liquid sample present is B = B app (1 + χ m ) , where B app is the magnetic feld in the ab- sence oF the liquid sample. Thus the magnetic susceptibility χ m is χ m = Δ B B app = 0 . 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 μ 0 . The permeability oF Free space is 1 . 25664 × 10 - 6 N / A 2 . I r cross sectional area is 2 . 141 cm 2 I o n c re , p e rm ea bil it y F 4 5 6 0 μ 7 t u s Fw ir i d Calculate the current necessary to produce a magnetic ±ux 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 . 0002141 m 2 , N = 700 turns , μ 0 =1 . 25664 × 10 - 6 N / A 2 , μ m = 4560 μ 0 , and Φ m . 00057 Wb . Basic concepts: Magnetic feld strength H = NI 2 π r . Magnetic ±ux density B = μ m H . Magnetic permeability μ m = 0 , where k = 1 + χ , and χ is the susceptibility.
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toupal (rgt374) – Homework 09 – Chiu – (58295) 2 Solution: To determine the magnetic feld strength H in Free space H = NI 2 π r , where r is radius oF toroid. Thus magnetic ±ux density B is B = μ m H = μ m 2 . Assuming no - B feld outside the solenoid and uniForm feld inside, the magnetic ±ux is Φ m = BA = μ m N I A 2 ThereFore, the required current is given by I = Φ m 2 μ m NA = Φ m 2 0 = (0 . 00057 Wb) 2 π (4560)(1 . 25664 × 10 - 6 N / A 2 ) × (0 . 1044 m) (700 turns)(0 . 0002141 m 2 ) = 0 . 435377 A . 004 10.0 points A square loop oF wire oF resistance R and side a is oriented with its plane perpendicular to a magnetic feld - B , as shown in the fgure. B B a I What must be the rate oF change oF the magnetic feld in order to produce a current I in the loop? 1. dB dt = I Ra 2. dt = IR a 2 correct 3. dt = Ra I 4. dt = Ia R 5. dt = 2 R Explanation: The emf produced is given by E = d Φ dt = a 2 dt . Using Ohm’s law we can solve For B = a 2 dt dt = a 2 .
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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

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