Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 67AP

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 67AP

The magnitude of the electric field in the wire is 8.00V/m in the positive x direction.

Explanation of Solution

Given information: Length of the cylindrical wire is 0.500m , diameter of the cylindrical wire is 0.200mm , resistivity of the wire is 4.00×108Ωm , potential maintained at left end of the wire at x=0 is 4.00V , potential maintained at the wire at x=0.500m is 0V .

Formula to calculate the magnitude of the electric field in the wire.

E=VLi^ (1)

Here,

E is the electric field in the wire.

V is the potential maintained at the wire.

L is the length of the cylindrical wire.

Substitute 4.00V for V , 0.500m for L in equation (1) to find E ,

E=4.00V0.500mi^=8.00x^V/m

Thus, the magnitude and direction of the electric field in the wire is 8.00V/m in the positive x direction.

Conclusion:

Therefore, the magnitude and direction of the electric field in the wire is 8.00V/m in the positive x direction.

(b)

To determine

The resistance of the wire.

(b)

Expert Solution
Check Mark

Answer to Problem 67AP

The resistance of the wire is 0.637Ω .

Explanation of Solution

Given information: Length of the cylindrical wire is 0.500m , diameter of the cylindrical wire is 0.200mm , resistivity of the wire is 4.00×108Ωm , potential maintained at left end of the wire at x=0 is 4.00V , potential maintained at the wire at x=0.500m is 0V .

Write the expression for the area of cross section of the cylindrical wire.

A=π4d2 (2)

Here,

A is the area of cross section of the cylindrical wire.

d is the diameter of the cylindrical wire.

Substitute 0.200mm for d in equation (2) to find A ,

A=π4(0.200mm×1m1000mm)2=3.14×108m2

Thus, the area of cross section of the cylindrical wire is 3.14×108m2 .

Formula to calculate the resistance of the wire.

R=ρLA (3)

Here,

R is the resistance of the wire.

L is the the length of the wire.

ρ is the resistivity of the wire.

Substitute 0.500m for L , 4.00×108Ωm for ρ , 3.14×108m2 for A in equation (3) to find R ,

R=(4.00×108Ωm)×0.500m3.14×108m2=0.6369Ω0.637Ω

Thus, the resistance of the wire is 0.637Ω .

Conclusion:

Therefore, the resistance of the wire is 0.637Ω .

(c)

To determine

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

(c)

Expert Solution
Check Mark

Answer to Problem 67AP

The magnitude and direction of the electric current in the wire is 6.28A in the positive x direction.

Explanation of Solution

Given information: Length of the cylindrical wire is 0.500m , diameter of the cylindrical wire is 0.200mm , resistivity of the wire is 4.00×108Ωm , potential maintained at left end of the wire at x=0 is 4.00V , potential maintained at the wire at x=0.500m is 0V .

Formula to calculate the magnitude of the electric current in the wire.

I=VRi^ (4)

Here,

I is the electric current in the wire.

V is the potential maintained at the wire.

L is the length of the cylindrical wire.

Substitute 4.00V for V , 0.637Ω for R in equation (4) to find I ,

I=4.00V0.637Ωi^=6.279i^A6.28i^A

Thus, the magnitude and direction of the electric current in the wire is 6.28A in the positive x direction.

Conclusion:

Therefore, the magnitude and direction of the electric current in the wire is 6.28A in the positive x direction.

(d)

To determine

The current density in the wire.

(d)

Expert Solution
Check Mark

Answer to Problem 67AP

The current density in the wire is 200MA/m2 .

Explanation of Solution

Given information: Length of the cylindrical wire is 0.500m , diameter of the cylindrical wire is 0.200mm , resistivity of the wire is 4.00×108Ωm , potential maintained at left end of the wire at x=0 is 4.00V , potential maintained at the wire at x=0.500m is 0V .

Formula to calculate the current density in the wire.

J=IA (5)

Here,

J is the current density in the wire.

Substitute 6.28A for I , 3.14×108m2 for A in equation (5) to find J ,

J=6.28A3.14×108m2=200000000A/m2=200MA/m2

Thus, the current density in the wire is 200MA/m2 .

Conclusion:

Therefore, the current density in the wire is 200MA/m2 .

(e)

To determine

To show: The expression for electric field in the wire is given by E=ρJ

(e)

Expert Solution
Check Mark

Answer to Problem 67AP

The expression for electric field in the wire is given by E=ρJ .

Explanation of Solution

Given information: Length of the cylindrical wire is 0.500m , diameter of the cylindrical wire is 0.200mm , resistivity of the wire is 4.00×108Ωm , potential maintained at left end of the wire at x=0 is 4.00V , potential maintained at the wire at x=0.500m is 0V .

From equation (1), write the expression for the electric field in the wire.

E=VLi^

Multiply by ρ in numerator and denominator in above expression.

E=ρ(VρL)i^ (6)

From equation (5), formula to calculate the current density in the wire.

J=IA

From equation (4), formula to calculate the magnitude of the electric current in the wire.

I=VRi^

Substitute VRi^ for I in equation (4) to find J ,

J=(VRi^)A=VRAi^ (7)

From equation (3), formula to calculate the resistance of the wire.

R=ρLA

Substitute ρLA for R in equation (7) to find J ,

J=V(ρLA)Ai^=VρLi^

Substitute J for VρLi^ in equation (6) to find E ,

E=ρJ

Thus, the expression for electric field in the wire is given by E=ρJ .

Conclusion:

Therefore, the expression for electric field in the wire is given by E=ρJ .

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Chapter 27 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 27 - Prob. 6OQCh. 27 - Prob. 7OQCh. 27 - Prob. 8OQCh. 27 - Prob. 9OQCh. 27 - Prob. 10OQCh. 27 - Prob. 11OQCh. 27 - Prob. 12OQCh. 27 - Prob. 13OQCh. 27 - Prob. 1CQCh. 27 - Prob. 2CQCh. 27 - Prob. 3CQCh. 27 - Prob. 4CQCh. 27 - Prob. 5CQCh. 27 - Prob. 6CQCh. 27 - Prob. 7CQCh. 27 - Prob. 8CQCh. 27 - Prob. 1PCh. 27 - A small sphere that carries a charge q is whirled...Ch. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - The quantity of charge q (in coulombs) that has...Ch. 27 - Prob. 10PCh. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - A wire 50.0 m long and 2.00 mm in diameter is...Ch. 27 - A 0.900-V potential difference is maintained...Ch. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - While taking photographs in Death Valley on a day...Ch. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - At what temperature will aluminum have a...Ch. 27 - Assume that global lightning on the Earth...Ch. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - The potential difference across a resting neuron...Ch. 27 - Prob. 41PCh. 27 - Prob. 42PCh. 27 - Prob. 43PCh. 27 - Prob. 44PCh. 27 - Prob. 45PCh. 27 - Prob. 46PCh. 27 - Prob. 47PCh. 27 - Prob. 48PCh. 27 - Prob. 49PCh. 27 - Prob. 50PCh. 27 - Prob. 51PCh. 27 - Prob. 52PCh. 27 - Prob. 53PCh. 27 - Prob. 54PCh. 27 - Prob. 55PCh. 27 - Prob. 56PCh. 27 - Prob. 57APCh. 27 - Prob. 58APCh. 27 - Prob. 59APCh. 27 - Prob. 60APCh. 27 - Prob. 61APCh. 27 - Prob. 62APCh. 27 - Prob. 63APCh. 27 - Review. An office worker uses an immersion heater...Ch. 27 - Prob. 65APCh. 27 - Prob. 66APCh. 27 - Prob. 67APCh. 27 - Prob. 68APCh. 27 - Prob. 69APCh. 27 - Prob. 70APCh. 27 - Prob. 71APCh. 27 - Prob. 72APCh. 27 - Prob. 73APCh. 27 - Prob. 74APCh. 27 - Prob. 75APCh. 27 - Prob. 76APCh. 27 - Review. A parallel-plate capacitor consists of...Ch. 27 - The dielectric material between the plates of a...Ch. 27 - Prob. 79APCh. 27 - Prob. 80APCh. 27 - Prob. 81APCh. 27 - Prob. 82CPCh. 27 - Prob. 83CPCh. 27 - Material with uniform resistivity is formed into...Ch. 27 - Prob. 85CP
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