Homework 6-solutions

# Homework 6-solutions - zhou(ejz224 – Homework 6 – shih...

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Unformatted text preview: zhou (ejz224) – Homework 6 – shih – (57480) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A total current of 90 mA flows through an infinitely long cylinderical conductor of radius 4 cm which has an infinitely long cylindrical hole through it of diameter r centered at r 2 along the x-axis as shown. x y What is the magnitude of the magnetic field at a distance of 15 cm along the positive x- axis? The permeability of free space is 4 π × 10 − 7 T · m / A . Assume the current density is constant throughout the conductor. Correct answer: 1 . 13846 × 10 − 7 T. Explanation: Basic Concepts: Magnetic Field due to a Long Cylinder B = μ I 2 π r . Principle of Superposition. Our goal is to model the given situation, which is complex and lacks symmetry, by adding together the fields from combinations of simpler current configurations which to- gether match the given current distribution. The combination of the currents in Fig. 2 will do so if we choose I cyl and I hole correctly. x y r I cyl = 4 3 I x y r 2 Hole I hole = − 1 3 I + Since the current is uniform, the current density J = I A is constant. Then J = I cyl A cyl = − I hole A hole Clearly, A cyl = π r 2 , and A hole = π r 2 4 , so I hole = − I cyl 4 . Note: The minus sign means I hole is flowing in the direction opposite I cyl and I , as it must if it is going to cancel with I cyl to model the hole. We also require I = I cyl + I hole . We then have I cyl = 4 3 I , and I hole = − 1 3 I . With these currents, the combination of the two cylinders in figure 2 gives the same net current and current distribution as the conductor in our problem. The magnetic fields are B cyl = μ parenleftbigg 4 3 I parenrightbigg 2 π x B hole = μ parenleftbigg − 1 3 I parenrightbigg 2 π ( x − r/ 2) , so the total magnetic field is B total = B cyl + B hole = μ I 6 π 4 x − 1 x − r 2 = μ I 6 π 3 x − 2 r x parenleftBig x − r 2 parenrightBig = (4 π × 10 − 7 T m / A) (90 mA) 6 π zhou (ejz224) – Homework 6 – shih – (57480) 2 × 3 (15 cm) − 2 (4 cm) (15 cm) parenleftbigg 15 cm − 4 cm 2 parenrightbigg = 1 . 13846 × 10 − 7 T . keywords: 002 (part 1 of 2) 10.0 points Consider two long parallel wires which are perpendicular to the plane of the paper ( i.e. , the x- y plane). Both wires carry the same current, I . Wire #1 intersects the plane a distance a above point O and wire #2 in- tersects the plane a distance a below point O . Point C is equidistant from both wires and is a distance a from point O ....
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## This note was uploaded on 04/09/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.

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Homework 6-solutions - zhou(ejz224 – Homework 6 – shih...

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