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

Question

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Answer 1

Question

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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)

To determine

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

(b)

Expert Solution

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)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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)

To determine

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

(d)

Expert Solution

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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Answer 2

Question

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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)

To determine

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

(b)

Expert Solution

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)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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)

To determine

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

(d)

Expert Solution

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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Answer 3

Question

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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)

To determine

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

(b)

Expert Solution

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)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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)

To determine

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

(d)

Expert Solution

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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Answer 4

Question

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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)

To determine

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

(b)

Expert Solution

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)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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)

To determine

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

(d)

Expert Solution

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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Answer 5

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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)

To determine

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

(b)

Expert Solution

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)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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)

To determine

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

(d)

Expert Solution

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 $0$ .

Answer 6

Chapter 19, Problem 59E

(a)

To determine

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

(a)

Expert Solution

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 $2k′Ira2$ .

Answer 7

Chapter 19, Problem 59E

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 59E

The magnetic field in region $r<a$ is $2k′Ira2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 $2k′Ira2$ .

Answer 13

(b)

To determine

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

(b)

Expert Solution

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 $2k′Ir$ .

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 59E

The magnetic field in region $a<r<b$ is $2k′Ir$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 $2k′Ir$ .

Answer 20

(c)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

(c)

Expert Solution

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 $2k′I(r2−b2)r(c2−b2)$ .

Answer 21

(c)

To determine

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

Answer 22

(c)

Expert Solution

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 $2k′I(r2−b2)r(c2−b2)$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

(d)

To determine

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

(d)

Expert Solution

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 $0$ .

Answer 27

(d)

To determine

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

Answer 28

(d)

Expert Solution

Answer to Problem 59E

The magnetic field in region $r>c$ is $0$ .

Answer 29

(d)

Expert Solution

Answer 30

(d)

Expert Solution

Answer 31

Expert Solution

Answer 32

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 $0$ .

Answer 33

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

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

Concept explainers

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

Answer to Problem 59E

Explanation of Solution

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

Answer to Problem 59E

Explanation of Solution

The magnetic field in region $b<r<c$ is $2k′I(r2−b2)r(c2−b2)$ .

Explanation of Solution

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

Answer to Problem 59E

Explanation of Solution

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Chapter 19 Solutions

The magnetic field for two conductors in region r &lt; a .

Concept explainers

The magnetic field for two conductors in region r<a<m:math><m:mrow><m:mi>r</m:mi><m:mo>&lt;</m:mo><m:mi>a</m:mi></m:mrow></m:math>.

Answer to Problem 59E

Explanation of Solution

The magnetic field for two conductors in region a<r<b<m:math><m:mrow><m:mi>a</m:mi><m:mo>&lt;</m:mo><m:mi>r</m:mi><m:mo>&lt;</m:mo><m:mi>b</m:mi></m:mrow></m:math>.

Answer to Problem 59E

Explanation of Solution

Explanation of Solution

The magnetic field for two conductors in region r>c<m:math><m:mrow><m:mi>r</m:mi><m:mo>&gt;</m:mo><m:mi>c</m:mi></m:mrow></m:math>.

Answer to Problem 59E

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions

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

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

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

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