47135_HW8

# 47135_HW8 - ragsdale(zdr82 HW8 ditmire(58335 This print-out...

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ragsdale (zdr82) – HW8 – ditmire – (58335) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A conductor consists oF an infnite number oF adjacent wires, each infnitely long and carrying a current I (whose direction is out-oF- the-page), thus Forming a conducting plane. A C IF there are n wires per unit length, what is the magnitude oF * B ? 1. B = μ 0 I 2 2. B =4 μ 0 I 3. B =2 μ 0 nI 4. B μ 0 5. B = μ 0 I 6. B μ 0 I 7. B = μ 0 4 8. B = μ 0 I 4 9. B = μ 0 2 correct 10. B = μ 0 Explanation: l W A C B B By symmetry the magnetic felds are equal and opposite through point A and C and hori- zontally oriented. ±ollowing the dashed curve in a counter-clockwise direction, we calculate ± * B · d*s , which by Ampere’s law is propor- tional to the current through the dashed loop coming out oF the plane oF the paper. In this problem this is a positive current. Hence * B along the horizontal legs points in the di- rection in which we Follow the dashed curve. Ampere’s Law is ± * B · d*s = μ 0 I. To evaluate this line integral, we use the rect- angular path shown in the fgure. The rectan- gle has dimensions l and w . The net current through the loop is nI l . Note that since there is no component oF * B in the direction oF w , we are only interested in the contributions along sides l ± * B · d*s Bl = μ 0 nl I B = μ 0 2 . 002 10.0 points A superconducting solenoid has 5810 turns / m and carries a current oF 2000 A. What is the magnetic feld generated inside the solenoid? The permeability oF Free space is 1 . 25664 × 10 - 6 T · m / A. Correct answer: 14 . 6021 T. Explanation: Let : n = 5810 turns / m , μ 0 =1 . 25664 × 10 - 6 T · m / A , and I = 2000 A .

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ragsdale (zdr82) – HW8 – ditmire – (58335) 2 The magnetic feld generated inside the solenoid is B = μ 0 nI = ( 1 . 25664 × 10 - 6 T · m / A ) × (5810 turns / m) (2000 A) = 14 . 6021 T . keywords: 003 (part 1 oF 2) 10.0 points A capacitor oF capacitance C has a charge Q at t = 0. At that time, a resistor oF resistance R is connected to the plates oF the charged capacitor. ±ind the magnitude oF the dis- placement current between the plates oF the capacitor as a Function oF time. 1. Q RC e - t/Q 2. RC Q e t/ ( RC ) 3. Q RC e - t/ ( RC ) correct 4. Q RC e t/ ( RC ) 5. RC Q e - t/ ( RC ) Explanation: Basic Concept RC circuits. Displacement Current. The displacement current is defned to be I d = ( 0 d Φ E dt . The electric feld inside a capacitor is essen- tially uniForm and E = q ( 0 A . Since the charge on a capacitor in a discharging RC circuit is given by q ( t )= Qe - t/RC , the displacement current is Found by I d = ( 0 d Φ E dt = ( 0 d dt ± q ( 0 A A ² = dq dt = - Q RC e - t/ ( RC ) . Note that the displacement current equals the actual current in the wires to the capacitor. Thus, the Ampere-Maxwell law tells us that * B will be the same regardless oF which current we evaluate. 004 (part 2 oF 2) 10.0 points Given C =3 μ ±, Q = 37 μ C, R = 384 kΩ, and ( 0 =8 . 85419 × 10 - 12 C 2 / N · m 2 , at what rate is the electric ²ux between the plates changing at time t =0 . 19 s?
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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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47135_HW8 - ragsdale(zdr82 HW8 ditmire(58335 This print-out...

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