Chapter 26, Problem 6P

Figure P26.6 represents a section of a conductor of nonuniform diameter carrying a current of I = 5.00 A. The radius of cross-section A₁ is r₁ = 0.400 cm. (a) What is the magnitude of the current density across A₁? The radius r₂ at A₂ is larger than the radius r₁ at A₁. (b) Is the current at A₂ larger, smaller, or the same? (c) Is the current density at A₂ larger, smaller, or the same? Assume A₂ = 4A₁. Specify the (d) radius, (e) current, and (f) current density at A₂.

Figure P26.6

To determine

The magnitude of the current density across $A_1$.

Answer to Problem 6P

The magnitude of the current density across $A_1$ is $99471.83\text{A}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$.

Write the expression for the area of cross section of a conductor.

$A_1=\pi r_1^2$ (1)

Here,

$A_1$ is the area of cross section of a conductor.

$r_1$ is the radius of cross section $A_1$.

Substitute $0.400\text{cm}$ for $r_1$ in equation (1) to find $A_1$,

$\begin{array}{c}{A}_1=\pi {\left(0.400\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =5.026\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

Thus, the area of cross section of a conductor is $5.026\times {10}^{-5}{\text{m}}^{\text{2}}$.

Formula to calculate the current density across $A_1$.

$J_1=\frac{I_1}{A_1}$ (2)

Here,

$J_1$ is the current density across $A_1$.

$I_1$ is the current carried by a conductor.

Substitute $5.00\text{A}$ for $I_1$, $5.026\times {10}^{-5}{\text{m}}^{\text{2}}$ for $A_1$ in equation (2) to find $J_1$,

$\begin{array}{c}{J}_1=\frac{5.00\text{A}}{5.026\times {10}^{-5}{\text{m}}^{\text{2}}}\\ =99471.83\text{A}/{\text{m}}^{\text{2}}\end{array}$

Thus, the magnitude of the current density across $A_1$ is $99471.83\text{A}/{\text{m}}^{\text{2}}$.

Conclusion:

Therefore, the magnitude of the current density across $A_1$ is $99471.83\text{A}/{\text{m}}^{\text{2}}$.

To determine

The reason that the current at $A_2$ is smaller, larger or remains same.

Answer to Problem 6P

The current at $A_2$ remains same because the current is not depending on cross sectional area. So, the current at $A_2$ is equal to the current at $A_1$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$.

The current is not depending on cross sectional area. So, the current at $A_2$ is equal to the current at $A_1$. Therefore, the current is the same.

Thus, the density at $A_2$ remains same because the current is not depending on cross sectional area. So, the current at $A_2$ is equal to the current at $A_1$.

Conclusion:

Therefore, the density at $A_2$ remains same because the current is not depending on cross sectional

area. So, the current at $A_2$ is equal to the current at $A_1$.

To determine

The reason that the current density at $A_2$ is smaller, larger or remains same.

Answer to Problem 6P

The current density at $A_2$ is smaller than the current density at $A_1$ because area of cross section at $A_2$ is greater than the area of cross section at $A_1$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$.

From equation (2),

Formula to calculate the current density across $A_1$.

$J_1=\frac{I_1}{A_1}$

Here,

$J_1$ is the current density across $A_1$.

Formula to calculate the current density across $A_2$.

$J_2=\frac{I_2}{A_2}$

Here,

$J_2$ is the current density across $A_2$.

$I_2$ is the current carried by a conductor.

From above relations, the current density is inversely proportional to area of cross section. From the figure given in the question, it is shown that:

$A_2>A_1$

Hence, the current density at $A_2$ is smaller than the current density at $A_1$.

$J_2>J_1$

Thus, the current density at $A_2$ is smaller than the current density at $A_1$ because area of cross section at $A_2$ is greater than the area of cross section at $A_1$.

Conclusion:

Therefore, the current density at $A_2$ is smaller than the current density at $A_1$ because area of cross section at $A_2$ is greater than the area of cross section at $A_1$.

To determine

The radius of cross section at $A_2$.

Answer to Problem 6P

The radius of cross section at $A_2$ is $\text{8}\text{.00}\text{mm}$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$, assume $A_2=4A_1$.

It is given that the expression for the crossectional area is:

$A_2=4A_1$ (3)

From equation (1),

Write the expression for the area of cross section of a conductor $A_1$.

$A_1=\pi r_1^2$

Write the expression for the area of cross section of a conductor $A_2$.

$A_2=\pi r_2^2$

Here,

$A_2$ is the area of cross section of a conductor.

$r_2$ is the radius of cross section $A_2$.

Substitute $\pi r_1^2$ for $r_1$, $\pi r_2^2$ for $A_2$ in equation (3) to find $r_2$,

$\begin{array}{c}\pi r_2^2=4\left(\pi r_1^2\right)\\ r_2^2=4r_1^2\\ r_2=\sqrt{4r_1^2}\\ =2r_1\end{array}$ (4)

Substitute $0.400\text{cm}$ for $r_1$ in equation (4) to find $r_2$,

Thus, the area of cross section of a conductor is $5.026\times {10}^{-5}{\text{m}}^{\text{2}}$.

$\begin{array}{c}{r}_2=2\times \left(0.400\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =8\times {10}^{-3}\text{m}\\ =\text{8}\text{.00}\text{mm}\end{array}$

Thus, the radius of cross section at $A_2$ is $\text{8}\text{.00}\text{mm}$.

Conclusion:

Therefore, the radius of cross section at $A_2$ is $\text{8}\text{.00}\text{mm}$.

To determine

The current for cross section at $A_2$.

Answer to Problem 6P

The current for cross section at $A_2$ is $5.00\text{A}$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$.

The current is not depending on cross sectional area. So, the current at $A_2$ is equal to the current at $A_1$. Therefore, the current is the same as for cross section 1.

$I_2=I_1=5.00\text{A}$

Thus, the current for cross section at $A_2$ is $5.00\text{A}$.

Conclusion:

Therefore, the current for cross section at $A_2$ is $5.00\text{A}$.

To determine

The magnitude of the current density across $A_2$.

Answer to Problem 6P

The magnitude of the current density across $A_2$ is $24875.62\text{A}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given information: Current carried by a conductor is $5.00\text{A}$, radius of cross section $A_1$ is $0.400\text{cm}$.

Write the expression for the area of cross section of a conductor $A_2$.

$A_2=\pi r_2^2$ (5)

Here,

$A_2$ is the area of cross section of a conductor.

$r_2$ is the radius of cross section $A_2$.

Substitute $\text{8}\text{.00}\text{mm}$ for $r_1$ in equation (5) to find $A_2$,

$\begin{array}{c}{A}_2=\pi {\left(\text{8}\text{.00}\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\\ =2.010\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Thus, the area of cross section of a conductor is $2.010\times {10}^{-4}{\text{m}}^{\text{2}}$.

Formula to calculate the current density across $A_2$.

$J_2=\frac{I_2}{A_2}$ (6)

Substitute $5.00\text{A}$ for $I_2$, $2.010\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A_2$ in equation (6) to find $J_2$,

$\begin{array}{c}{J}_2=\frac{5.00\text{A}}{2.010\times {10}^{-4}{\text{m}}^{\text{2}}}\\ =24875.62\text{A}/{\text{m}}^{\text{2}}\end{array}$

Thus, the magnitude of the current density across $A_2$ is $24875.62\text{A}/{\text{m}}^{\text{2}}$.

Conclusion:

Therefore, the magnitude of the current density across $A_2$ is $24875.62\text{A}/{\text{m}}^{\text{2}}$.