Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
15th Edition
ISBN: 9781305289963
Author: Debora M. Katz
Publisher: Cengage Custom Learning
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Chapter 28, Problem 82PQ

A conducting material with resistivity ρ is shaped into a wire of length ( with a tapered cross section that decreases from radius r1 on the left end to r2 on its right end (Fig. F28.82). If the current density in the wire is a constant as a function of the horizontal distance x along the wire, what is the resistance of this tapered wire?

Chapter 28, Problem 82PQ, A conducting material with resistivity  is shaped into a wire of length ( with a tapered cross

Expert Solution & Answer
Check Mark
To determine

The resistance of the tapered wire.

Answer to Problem 82PQ

The resistance of the tapered wire is R=ρπ(lr1r2)_.

Explanation of Solution

The diagram of the tapered wire is given in Figure P28.82. The length of the wire is l, the radii of the two end cross sections of the wire is r1 and r2 (r1>r2). The horizontal distance along the wire is represented as x. The current density in the wire is constant as a function of x.

From the geometry of the longitudinal section of the wire, it can be deduced that,

  (r1r)x=(r1r2)l                                                                                                  (I)

Here, r is the radius at a distance x.

From equation (I), write the expression for the radius at a distance x from the left end of the wire.

  r=r1(r1r2)xl                                                                                                  (II)

Write the general expression for the resistance of the wire.

  R=ρlA                                                                                                               (III)

Here, R is the resistance, ρ is the resistivity, and A is the cross-sectional area of the wire.

Since the cross-sectional area of the wire changes as the distance x changes, for a disk shaped element of this wire with radius r and thickness dx, equation (II) can be rewritten as,

  dR=ρdxπr2                                                                                                         (IV)

Use equation (II) in (IV).

  dR=ρπdx[r1(r1r2)(xl)]2                                                                                  (V)

Integrate equation (V) with the limits x=0 to x=l to find the resistance of the wire.

  R=ρπ0ldx[r1(r1r2)(xl)]2=ρπ0ldx[(r2r1l)x+r1]2                                                                                (VI)

Take y=(r2r1l)x+r1, so that dy=(r2r1l)dx. Thus, dx=dy(r2r1l). Use this in the integral (VI)

  R=ρπ1(r2r1l)dyy2=ρπ1(r2r1l)(1y)                                                                                          (VII)

Use y=(r2r1l)x+r1 in (VII) and apply the limits.

  R=(ρπ)[1(r2r1l)[(r2r1l)x+r1]]|x=0x=l=(ρπ)[1[(r2r1l)2x+(r2r1l)r1]]|x=0x=l=(ρπ)[1[(r2r1l)2l+(r2r1l)r1]1(r2r1l)r1]=(ρπ)[1[(r2r1)2l+(r2r1l)r1]1(r2r1l)r1]                                        (VIII)

Reduce equation (VIII).

  R=(ρπ)1(r2r1l)(1r2r1+r11r1)=(ρπ)1(r2r1l)(1r21r1)=(ρπ)(lr2r1)(r2r1r1r2)=ρπ(lr1r2)

Conclusion:

Therefore, the resistance of the tapered wire is R=ρπ(lr1r2)_.

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Chapter 28 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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