Question

A cylindrical resistor of radius 5.00mm and length2.0 cmis made of material that has a resistivity of $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\Omega }\mathbf{.}\mathbf{m}$ .What are (a) the magnitude of the current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0 W?

Short answer

1. The magnitude of the current density is $1.3\times {10}^5\mathrm{A}/{\mathrm{m}}^2$.
2. The potential difference when the energy dissipation rate in the resistor is 1.0 W is $9.4\times {10}^{-2}\mathrm{V}$.

Step-by-step solution

The given data:

The radius of the cylindrical resistor,$\mathrm{r}=5\mathrm{mm}=5\times {10}^{-3}\mathrm{m}$

Length of the resistor,$\mathrm{L}=2\mathrm{cm}=2\times {10}^{-5}\mathrm{m}$

The resistivity of the material, $\mathrm{p}=3.5\times {10}^{-5}\mathrm{\Omega }.\mathrm{m}$

Energy dissipation rate, P = 1.0 W

Understanding the concept of the current density and potential difference:

Using the current density formula, you first compare current density and power; then find the total current density. From Equations 26-26, you can conclude that power is the product of current and potential, where current can be written in the form of current. So then calculate the total potential difference.

Formulas:

The current density of a material having uniform current flow,

$\mathrm{J}=\frac{\mathrm{i}}{\mathrm{A}}$ ….. (1)

Here, i is the current and A is the area.

The electric power in terms of potential difference,

P = iV ….. (2)

Here, V is the potential difference.

The rate of energy dissipation from the system,

$\mathrm{P}={\mathrm{i}}^2\mathrm{R}$ ….. (3)

The resistance of a material,

$\mathrm{R}=\frac{\mathrm{pL}}{\mathrm{A}}$ ….. (4)

Here, p is the resistivity and L is the length.

The cross-sectional area of a circular surface,

$\mathrm{A}={\mathrm{\pi r}}^2$ ….. (5)

(a) Calculation of the magnitude of the current density:

Substituting the value of current from equation (1) and the value of resistance from equation (4) in equation (3), you can get the following equation of the power generated as follows:

$$\begin{gathered} \mathrm{P}={\mathrm{J}}^2{\mathrm{A}}^2\frac{\mathrm{pL}}{\mathrm{A}} \\ ={\mathrm{J}}^2\mathrm{ApL} \end{gathered}$$

Rearrange the above equation for current density and substituting the value from equation (5), you have the magnitude of the current density using the given data as follows:

$$\begin{gathered} \mathrm{J}=\sqrt{\frac{\mathrm{P}}{{\mathrm{\pi r}}^2\mathrm{pL}}} \\ =\sqrt{\frac{1}{3.14\times {\left(5\times {10}^{-3}\right)}^2\times 3.5\times {10}^{-5}\times 2\times {10}^{-2}}} \\ =\sqrt{\frac{1}{549.5\times {10}^{-13}}} \\ =\sqrt{1.8198\times {10}^{10}} \\ \mathrm{J}=1.349\times {10}^5 \\ =1.3\times {10}^5\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the current density is $1.3\times {10}^5\mathrm{A}/{\mathrm{m}}^2$.

(b) Calculation of the potential difference:

Now, substituting the value of current of equation (1) in equation (2), you can get the equation of the rate of dissipated energy as follows:

P = JAV

By rearranging the above equation and using area value of equation (5), you obtain the value of the potential difference using the given data as follows:

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{P}}{{\mathrm{J\pi r}}^2} \\ =\frac{1}{1.349\times {10}^5\times 3.14\times {\left(5\times {10}^{-3}\right)}^2} \\ =9.4\times {10}^{-2}\mathrm{V} \end{gathered}$$

Hence, the value of the potential difference is $9.4\times {10}^{-2}\mathrm{V}$.