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 47PQ

(a)

To determine

The magnitude and direction of the electric field in the wire.

(a)

Expert Solution
Check Mark

Answer to Problem 47PQ

The magnitude and direction of the electric field in the wire are 4.27V/m and from the left.

Explanation of Solution

Write the relation for the electric field in the wire.

   Ex=dVdx=ΔVΔx                                                                                                    (1)

Here, Ex is electric field in the wire, ΔV/Δx is the slope of the potential which is  from high to low potential points.

Conclusion:

Substitute (03.20V) for ΔV and 0.750m for Δx to find Ex.

  Ex=(03.20V)0.750m=4.27V/mfromlefttoright

Therefore, the magnitude and direction of the electric field in the wire are 4.27V/m and from the left.

(b)

To determine

The resistance of the wire.

(b)

Expert Solution
Check Mark

Answer to Problem 47PQ

The resistance of the wire is 0.168Ω.

Explanation of Solution

Write the relation for the resistance of the cylindrical silver wire.

   R=ρlA=ρlπr2                                                                                                     (2)

Here, R is resistance of the wire, l is length of the wire, and A is area of the wire, and ρ is resistivity.

Conclusion:

Substitute 1.586×108Ωm for ρ, 0.750m for l, and 0.150×103m for r to find R in above equation.

   R=(1.586×108Ωm)0.750mπ(0.150×103m)2=0.168Ω

The resistance of the cylindrical silver wire is 0.168Ω.

(c)

To determine

The magnitude and direction of the current in the wire.

(c)

Expert Solution
Check Mark

Answer to Problem 47PQ

The magnitude and direction of the current in the wire are 19.0A fromlefttoright.

Explanation of Solution

Write the relation for the current in the cylindrical silver wire.

   I=ΔVR                                                                                                    (2)

Here, I is the current in the wire and ΔV is potential difference.

Conclusion:

Substitute 3.20V for ΔV and 0.168Ω for R to find I.

   I=3.20V0.168Ω=19.0A

The current is in positive sign represents that it flows from left to right.

The magnitude and direction of the current in the wire are 19.0A fromlefttoright.

(d)

To determine

The current density in the wire.

(d)

Expert Solution
Check Mark

Answer to Problem 47PQ

The current density in the wire is 2.69×108A/m3.

Explanation of Solution

Write the relation for the current density in the wire.

   J=IA=Iπr2                                                                                                        (3)

Here, J is current density.

Conclusion:

Substitute 19.0A for I and 0.150×103m for r in equation (3) to find J.

   J=19.0A0.150×103m=2.69×108A/m2

Therefore, the current density in the wire is 2.69×108A/m3.

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Chapter 28 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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