General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 19, Problem 59E

(a)

To determine

The magnetic field for two conductors in region r<a.

(a)

Expert Solution
Check Mark

Answer to Problem 59E

The magnetic field in region r<a is 2kIra2.

Explanation of Solution

Write the expression for Ampere’s law.

    Bdl=μ0I        (1)

Here, B is the magnetic field, dl is an elemental length, μ0 is the free space permeability and I is the current.

Write the expression for current density.

    J=IA        (2)

Here, J is the current density and A is the surface area.

Conclusion:

In the region r<a coaxial cable is only current contributor

Substitute (πa2) for A in equation (2).

    J=Iπa2

Here, a is the radius of the cable.

Substitute I for I and (πr2) for A in equation 2).

    J=Iπr2

Here, I is the current in region r<a and r is the distance of a point in that region.

Substitute Iπa2 for J in the above equation.

    Iπa2=Iπr2

Rearrange the above equation.

    I=Iπr2πa2=Ir2a2

Substitute Ir2a2 for I in equation (1).

    Bdl=μ0Ir2a2

Substitute (2πr) for the close integral in the above equation.

    B(2πr)=μ04π(4π)Ir2a2

Substitute k for μ04π in the above expression.

    B=2kIra2

Thus, the magnetic field in region r<a is 2kIra2.

(b)

To determine

The magnetic field for two conductors in region a<r<b.

(b)

Expert Solution
Check Mark

Answer to Problem 59E

The magnetic field in region a<r<b is 2kIr.

Explanation of Solution

Write the expression for Ampere’s law.

    Bdl=μ0I

Conclusion:

Rearrange equation (1).

    Bdl=μ0I

Substitute (2πr) for the close integral and 4πk for μ0 in the above equation.

    B(2πr)=(4πk)IB=2kIr

Here, k is a constant and r is the distance of a point in that region.

Thus, the magnetic field in region a<r<b is 2kIr.

(c)

To determine

The magnetic field in region b<r<c is 2kI(r2b2)r(c2b2).

(c)

Expert Solution
Check Mark

Explanation of Solution

Write the expression for Ampere’s law.

    Bdl=μ0I

Write the expression for current density.

    J=IA

Conclusion:

In the region b<r<c tube is only current contributor

Substitute π(c2b2) for A in equation (2).

    J=Iπ(c2b2)

Here, b is the inner radius of the tube and c is the outer radius of the tube.

Substitute I for I and π(r2b2) for A in equation 2).

    J=Iπ(r2b2)

Here, I is the current in region b<r<c and r is the distance of a point in that region.

Substitute Iπ(c2b2) for J in the above equation.

    Iπ(c2b2)=Iπ(r2b2)

Rearrange the above equation.

    I=Iπ(r2b2)π(c2b2)=I(r2b2)(c2b2)

Substitute I(r2b2)(c2b2) for I in equation (1).

    Bdl=μ0I(r2b2)(c2b2)

Substitute (2πr) for the close integral in the above equation.

    B(2πr)=μ04π(4π)I(r2b2)(c2b2)

Substitute k for μ04π and solve the above expression.

    B=2kI(r2b2)r(c2b2)

Thus, the magnetic field in region b<r<c is 2kI(r2b2)r(c2b2).

(d)

To determine

The magnetic field for two conductors in region r>c.

(d)

Expert Solution
Check Mark

Answer to Problem 59E

The magnetic field in region r>c is 0.

Explanation of Solution

Write the expression for Ampere’s law.

    Bdl=μ0I

Conclusion:

In the region r>c there is no contribution for current.

Substitute 0 for the I in equation (1).

    Bdl=μ0(0)B=0

Thus, the magnetic field in region r>c is 0.

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