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Unformatted text preview: zhou (ejz224) – Homework 6 – shih – (57480) 1 This printout should have 17 questions. Multiplechoice 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 xaxis 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.
 Spring '08
 morrison
 Physics, Current, Work

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