A copper wire of cross-sectional area $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathbf{m}}^{\mathbf{2}}$and length 4.00mhas a current of 2.00Auniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in 30min?

1. The cross-sectional area of wire is$\mathrm{A}=2.00\times {10}^{-6}{\mathrm{m}}^2$
2. The length of the wire is L = 4.00m
3. The current through the wire is i = 2.00 A

The electric field is equal to the product of resistivity and current density of the current in the material. The rate of energy dissipation can be written as the product of the square of current and resistance.

By using equations 26-11, 26-16, and 26-27, we can find the magnitude of the electric field along the wire, the resistance, and the power dissipated respectively. Then, by assuming a steady rate and using the value of time, we can find the amount of thermal energy generated in 30min.

Formulae:

The electric field due to the current density,$\overrightarrow{E}=p\overrightarrow{J}$ …(i)

Here,$\overrightarrow{E}$ is the electric field, p is the resistivity, $\overrightarrow{J}$and is the current density.

The current density due to the current flow through the area,$\left|\overrightarrow{J}\right|=\frac{i}{A}$ …(ii)

Here,$\overrightarrow{J}$ is the current density, i is current, A is the area of cross-section.

The resistance of the material,$R=p\frac{L}{A}$ …(iii)

Here, R is resistance, L is the length of the wire, p is the resistivity, A is the area of cross-section.

The power generated due to energy transfer,$P=i^{2}R$ …(iv)

Here, P is the power, i is current, R is the resistance.

The energy transferred from one system to another, E = Pt …(v)

Here, P is the power, E is electric energy, t is the time.

We can get the magnitude of the electric field along the wire using the given data and substituting equation (ii) in equation (i) as follows:

$$\begin{gathered} \left|\overrightarrow{E}\right|=p\left(\frac{i}{A}\right) \\ =1.69\times {10}^{-8}\mathrm{\Omega m}\times \left(\frac{2.00}{2.00\times {10}^{-6}{\mathrm{m}}^2}\right) \\ =1.69\times {10}^{-2}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the value of the field is $1.69\times {10}^{-2}\mathrm{V}/\mathrm{m}$.

At first, we can get the resistance of the wire using the given data in equation (iii) as follows:

$$\begin{gathered} R=1.69\times {10}^{-8}\mathrm{\Omega m}\times \left(\frac{4.0}{2.00\times {10}^{-6}{\mathrm{m}}^2}\right) \\ =0.0338\mathrm{\Omega } \end{gathered}$$

Now, using this above value, we can get the value of the power in equation (iv) as follows:

$$\begin{gathered} P={(2.00A)}^2\times 0.0338\Omega \\ =0.135\mathrm{W} \end{gathered}$$

Assuming a steady rate, the amount of thermal energy generated in 30min is found using equating (v) as follows:

$$\begin{gathered} E=0.135W\times (30\times 60)\mathrm{s} \\ =2,43\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the amount of generated thermal energy is $2.43\times {10}^2\mathrm{J}$.