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 32, Problem 60AP

(a)

To determine

Whether the force on one plate can be accounted for by thinking of the electric filed between the plates as exerting a negative pressure equal to the energy density of the electric field.

(a)

Expert Solution
Check Mark

Answer to Problem 60AP

The force on one plate can be accounted for by thinking of the electric filed between the plates as exerting a negative pressure equal to the energy density of the electric field.

Explanation of Solution

Write the expression to calculate the electric fields between the plates.

    E=σε0

Here, E is the electric field, σ is the stress and ε0 is the permittivity of free space.

Write the expression to calculate the force.

    F=ε0E2

Here, F is the force.

Write the expression to calculate the force on one plate.

    F'=F2F'=ε0E22                                                                                                                  (I)

Write the expression to calculate the energy density of the electric field.

    UE=12ε0E2                                                                                                           (II)

Here, UE is the energy density.

Negative sign indicates that the plates are drawn towards each other.

Equate equation (I) and (II).

    F'=UEε0E22=ε0E22

Conclusion:

Therefore, the force on one plate can be accounted for by thinking of the electric filed between the plates as exerting a negative pressure equal to the energy density of the electric field.

(b)

To determine

The force per area acting on the sheet due to the magnetic field.

(b)

Expert Solution
Check Mark

Answer to Problem 60AP

The force per area acting on the sheet due to the magnetic field is (μ0Jx22).

Explanation of Solution

Write the expression for magnetic field due to one plate using Ampere’s law.

    B=μ0Jx2

Here, B is the magnetic field, μ0 is the permeability of free space and Jx is the current density.

Write the expression to calculate the force per unit area.

    F=BJx

Here, F is the force.

Substitute (μ0Jx2) for B in the above equation to calculate F.

    F=(μ0Jx2)Jx=μ0Jx22

Conclusion:

Therefore, the force per area acting on the sheet due to the magnetic field is (μ0Jx22).

(c)

To determine

The net magnitude field between the sheets and the field outside of the volume between them.

(c)

Expert Solution
Check Mark

Answer to Problem 60AP

The net magnitude field between the sheets is (μ0Jx)  and the field outside of the volume is 0.

Explanation of Solution

Write the expression for magnetic field due to one plate using Ampere’s law.

    B=μ0Jx2

Here, B is the magnetic field, μ0 is the permeability of free space and Jx is the current density.

Write the expression to calculate the total magnetic field.

    B'=2B

Substitute (μ0Jx2) for B in the above equation to calculate B'.

    B'=2(μ0Jx2)=μ0Jx

The magnetic field between the sheets is perpendicular to the plane of the page and magnetic field outside is zero.

Conclusion:

Therefore, the net magnitude field between the sheets is (μ0Jx)  and the field outside of the volume is 0.

(d)

To determine

The energy density in the magnetic field between the sheets.

(d)

Expert Solution
Check Mark

Answer to Problem 60AP

The energy density in the magnetic field between the sheets is μ02Jx22.

Explanation of Solution

Write the expression for energy density in the magnetic field.

    Ed=B22μ0

Substitute μ0Jx for B in the above equation.

    Ed=(μ0Jx)22=μ02Jx22

Conclusion:

Therefore, the energy density in the magnetic field between the sheets is μ02Jx22.

(e)

To determine

The force on one sheet can be accounted by magnetic field between the sheets as exerting a positive pressure equal to its energy density.

(e)

Expert Solution
Check Mark

Answer to Problem 60AP

The force on one sheet is be accounted by magnetic field between the sheets as exerting a positive pressure equal to its energy density.

Explanation of Solution

Write the expression total magnetic field.

    B'=μ0JxJx=B'μ0

Write the expression for force on one sheet of paper.

    F=B×Jx

Substitute (B'μ0) for Jx and (B'2) for B in the above equation to calculate F.

    F=B'2(B'μ0)F=12μ0B'2                                                                                                          (III)

Write the expression for energy density.

    UB=12μ0B'2                                                                                                          (IV)

Here, UB is the energy density.

Conclusion:

Therefore, the force on one sheet is be accounted by magnetic field between the sheets as exerting a positive pressure equal to its energy density.

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Chapter 32 Solutions

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

Ch. 32 - Prob. 6OQCh. 32 - Prob. 7OQCh. 32 - Prob. 1CQCh. 32 - Prob. 2CQCh. 32 - Prob. 3CQCh. 32 - Prob. 4CQCh. 32 - Prob. 5CQCh. 32 - Prob. 6CQCh. 32 - The open switch in Figure CQ32.7 is thrown closed...Ch. 32 - Prob. 8CQCh. 32 - Prob. 9CQCh. 32 - Prob. 10CQCh. 32 - Prob. 1PCh. 32 - Prob. 2PCh. 32 - Prob. 3PCh. 32 - Prob. 4PCh. 32 - Prob. 5PCh. 32 - Prob. 6PCh. 32 - Prob. 7PCh. 32 - Prob. 8PCh. 32 - Prob. 9PCh. 32 - Prob. 10PCh. 32 - Prob. 11PCh. 32 - Prob. 12PCh. 32 - Prob. 13PCh. 32 - Prob. 14PCh. 32 - Prob. 15PCh. 32 - Prob. 16PCh. 32 - Prob. 17PCh. 32 - Prob. 18PCh. 32 - Prob. 19PCh. 32 - Prob. 20PCh. 32 - Prob. 21PCh. 32 - Prob. 22PCh. 32 - Prob. 23PCh. 32 - Prob. 24PCh. 32 - Prob. 25PCh. 32 - Prob. 26PCh. 32 - Prob. 27PCh. 32 - Prob. 28PCh. 32 - Prob. 29PCh. 32 - Prob. 30PCh. 32 - Prob. 31PCh. 32 - Prob. 32PCh. 32 - Prob. 33PCh. 32 - Prob. 34PCh. 32 - Prob. 35PCh. 32 - Prob. 36PCh. 32 - Prob. 37PCh. 32 - Prob. 38PCh. 32 - Prob. 39PCh. 32 - Prob. 40PCh. 32 - Prob. 41PCh. 32 - Prob. 42PCh. 32 - Prob. 43PCh. 32 - Prob. 44PCh. 32 - Prob. 45PCh. 32 - Prob. 46PCh. 32 - Prob. 47PCh. 32 - Prob. 48PCh. 32 - Prob. 49PCh. 32 - Prob. 50PCh. 32 - Prob. 51PCh. 32 - Prob. 52PCh. 32 - Prob. 53PCh. 32 - Prob. 54PCh. 32 - Prob. 55PCh. 32 - Prob. 56PCh. 32 - Prob. 57PCh. 32 - Prob. 58PCh. 32 - Electrical oscillations are initiated in a series...Ch. 32 - Prob. 60APCh. 32 - Prob. 61APCh. 32 - Prob. 62APCh. 32 - A capacitor in a series LC circuit has an initial...Ch. 32 - Prob. 64APCh. 32 - Prob. 65APCh. 32 - At the moment t = 0, a 24.0-V battery is connected...Ch. 32 - Prob. 67APCh. 32 - Prob. 68APCh. 32 - Prob. 69APCh. 32 - Prob. 70APCh. 32 - Prob. 71APCh. 32 - Prob. 72APCh. 32 - Prob. 73APCh. 32 - Prob. 74APCh. 32 - Prob. 75APCh. 32 - Prob. 76APCh. 32 - Prob. 77APCh. 32 - Prob. 78CPCh. 32 - Prob. 79CPCh. 32 - Prob. 80CPCh. 32 - Prob. 81CPCh. 32 - Prob. 82CPCh. 32 - Prob. 83CP
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