91影视

The given data can be given below as:

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

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,

$i=鈭�\left|\stackrel{鈫�}{J}\right|dA$ 鈥� (i)

The power dissipated during the energy transfer is,

$P=iV$ 鈥� (ii)

The thermal energy generated during the transfer is,

$Q=P鈭�t$ 鈥� (iii)

The cross-sectional area of the wire is,

$A={\mathrm{蟺谤}}^2$ 鈥� (iv)

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

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

Here, the expression of $\left|\stackrel{鈫�}{J}\right|=k{r}^{2}anddA=2\mathrm{蟺谤}$. The value of k is $\left(2.75脳{10}^{10}A/m^4\right)$.

Substitute the values in the above equation.

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

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

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

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

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

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