Q52P - Fundamentals Of Physics

Question

The current-density magnitude in a certain circular wire is $\mathbf{J}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{4}}\mathbf{)}{\mathbf{r}}^{\mathbf{2}}$, where r is the radial distance out to the wire’s radius of 3.00 mm. The potential applied to the wire (end to end) is 6.00 V. How much energy is converted to thermal energy in 1.00 h?

Step-by-Step Solution

Answer

The energy converted to thermal energy during time is $1.00\mathrm{h} \mathrm{is} 7.56\times {10}^5 \mathrm{J}$.

Step 1: Identification of the given data

The given data can be given below as:

• The current density magnitude is $\left|\overrightarrow{J}\right|=\left(2.75\times {10}^{10}\mathrm{A}/{\mathrm{m}}^4\right){\mathrm{r}}^2$
• The radius of wire is $R=3.00 \mathrm{mm}=3.00\times {10}^{-3}\mathrm{m}$.
• The potential is V = 60.0 V.
• The time is $∆t=1h=3600 s$.

Step 2: Understanding the concept of the flow of current, energy and power

In the problem, by using (i), we can find the current flowing through the wire. Then, we can use this value of current to find power using (ii). And last, using this value of power, we can find the thermal energy generated in 1.00 h.

 

Formulae:

The current density of the current flowing through the area,

$i=\int \left|\overrightarrow{J}\right|dA$                                                                                                                    … (i)

 

The power dissipated during the energy transfer is,

 $P=iV$                                                                                                                       … (ii)

 

The thermal energy generated during the transfer is,

$Q=P∆t$                                                                                                                    … (iii)

 

The cross-sectional area of the wire is, 

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

Step 3: Calculation of the converted energy

Using the area formula of equation (iv) in equation (i), we can get the value of the current flowing along the wire as follows: 

$$\begin{gathered} i={\int }_0^R\left(k{r}^2\right)\left(2\mathrm{\pi r}\right)dr \\ =\frac{1}{2}k{\mathrm{\pi R}}^4 \end{gathered}$$

Here, the expression of $\left|\overrightarrow{J}\right|=k{r}^2 and dA=2\mathrm{\pi r}$. The value of k is $\left(2.75\times {10}^{10}A/m^4\right)$.

 

Substitute the values in the above equation.

$$\begin{gathered} i=\frac{1}{2}\times 2.75\times {10}^{10}A/m^4\times \mathrm{\pi }\times {\left(3.00\times {10}^{-3}\mathrm{m}\right)}^4 \\ =3.50 \mathrm{A} \end{gathered}$$

 

Using this above value in equation (ii), we can get the power dissipated as follows:

$$\begin{gathered} P=3.50 A\times 60.0 V \\ =210 \mathrm{W} \end{gathered}$$

 

Now, the thermal energy generated in can be calculated using the given data in equation (iii) as follows:

$$\begin{gathered} Q=210 \mathrm{W}\times 3600 \mathrm{s}\left(\frac{1 \mathrm{J}}{1 \mathrm{W}.\mathrm{s}}\right) \\ =7.56\times {10}^5 \mathrm{J} \end{gathered}$$

 

Hence, the value of the thermal energy is $7.56\times {10}^5 \mathrm{J}$.