
Concept explainers
(a)
The magnetic field for two conductors in region r<a.
(a)

Answer to Problem 59E
The magnetic field in region r<a is 2k′Ira2.
Explanation of Solution
Write the expression for Ampere’s law.
∮→B⋅→dl=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=I′πr2
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=I′πr2
Rearrange the above equation.
I′=Iπr2πa2=Ir2a2
Substitute Ir2a2 for I in equation (1).
B∮dl=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=2k′Ira2
Thus, the magnetic field in region r<a is 2k′Ira2.
(b)
The magnetic field for two conductors in region a<r<b.
(b)

Answer to Problem 59E
The magnetic field in region a<r<b is 2k′Ir.
Explanation of Solution
Write the expression for Ampere’s law.
∮→B⋅→dl=μ0I
Conclusion:
Rearrange equation (1).
B∮dl=μ0I
Substitute (2πr) for the close integral and 4πk′ for μ0 in the above equation.
B(2πr)=(4πk′)IB=2k′Ir
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 2k′Ir.
(c)
The magnetic field in region b<r<c is 2k′I(r2−b2)r(c2−b2).
(c)

Explanation of Solution
Write the expression for Ampere’s law.
∮→B⋅→dl=μ0I
Write the expression for current density.
J=IA
Conclusion:
In the region b<r<c tube is only current contributor
Substitute π(c2−b2) for A in equation (2).
J=Iπ(c2−b2)
Here, b is the inner radius of the tube and c is the outer radius of the tube.
Substitute I′ for I and π(r2−b2) for A in equation 2).
J=I′π(r2−b2)
Here, I′ is the current in region b<r<c and r is the distance of a point in that region.
Substitute Iπ(c2−b2) for J in the above equation.
Iπ(c2−b2)=I′π(r2−b2)
Rearrange the above equation.
I′=Iπ(r2−b2)π(c2−b2)=I(r2−b2)(c2−b2)
Substitute I(r2−b2)(c2−b2) for I in equation (1).
B∮dl=μ0I(r2−b2)(c2−b2)
Substitute (2πr) for the close integral in the above equation.
B(2πr)=μ04π(4π)I(r2−b2)(c2−b2)
Substitute k′ for μ04π and solve the above expression.
B=2k′I(r2−b2)r(c2−b2)
Thus, the magnetic field in region b<r<c is 2k′I(r2−b2)r(c2−b2).
(d)
The magnetic field for two conductors in region r>c.
(d)

Answer to Problem 59E
The magnetic field in region r>c is 0.
Explanation of Solution
Write the expression for Ampere’s law.
∮→B⋅→dl=μ0I
Conclusion:
In the region r>c there is no contribution for current.
Substitute 0 for the I in equation (1).
∮→B⋅→dl=μ0(0)B=0
Thus, the magnetic field in region r>c is 0.
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