Question

A potential difference of 1.20 Vwill be applied to a 33.0 mlength of 18-gauge copper wire (diameter = 0.0400 in). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

Short answer

1. The current in the wire is 1.74 A.
2. The magnitude of the current density is $2.15\times {10}^6\mathrm{A}/{\mathrm{m}}^2$.
3. The magnitude of the electric field within the wire is$3.63\times {10}^{-2}\mathrm{V}/\mathrm{m}$.
4. The rate at which the thermal energy will appear in the wire is 2.09 W.

Step-by-step solution

Significance of the given data

The given data can be expressed below as:

• The potential difference is V = 1.20 V.
• The length of copper wire is L = 33.0 m.
• The diameter of copper wire is $\mathrm{d}=0.0400\mathrm{in}=0.1016\times {10}^{-2}\mathrm{m}$.

 Step 2: Understanding the concept of the flow of current and field

By using equation (i), we can find the current through the copper wire. By using this value of current in the formula for the magnitude of the current density vector, we can find the magnitude of the current density.

By using the formula for the magnitude of the electric field, we can find its value. By using the value of current in equation (vi), we can find the rate at which the thermal energy will appear in the wire.

Formulae:

The current equation from Ohm’s law is,

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}$ … (i)

The resistance of the material is,

$R=p\frac{L}{A}$ … (ii)

The cross-sectional area in terms of diameter is,

$A=\frac{\pi d^2}{4}$ … (iii)

The magnitude of the current density vector is,

role="math" localid="1661831398240" $\left|\overrightarrow{J}\right|\frac{i}{A}$ … (iv)

The magnitude of the electric field for potential difference is,

$E=\frac{V}{L}$ … (v)

The rate at which the thermal energy will appear in the wire is,

P = iV … (vi)

(a) Calculation of the current in the wire

Substituting the value of area from equation (iii) in the equation (ii), we can get the value of the current from equation (i) in the wire as follows:

$i=\frac{\pi d^{2}V}{4pL}$

Here,is the density of copper whose value is.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{i}=\left[\frac{\mathrm{\pi }\times \left(0.1016\times {10}^{-2}\mathrm{m}\right)\times 1.20\mathrm{V}}{4\times \left(1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}\right)\times 33.0\mathrm{m}}\right] \\ =1.74\mathrm{A} \end{gathered}$$

Hence, the value of the current is 1.74 A.

(b) Calculation of the magnitude of the current density

The magnitude of the current density vector can be calculated using equation (iii) in equation (iv) as follows:

$\left|\overrightarrow{J}\right|\frac{4i}{\pi d^2}$

Substitute the values in the above equation.

$$\begin{gathered} \left|\overrightarrow{\mathrm{J}}\right|=\frac{4\times 1.74\mathrm{A}}{\mathrm{\pi }\times {\left(0.1016\times {10}^{-2}\mathrm{m}\right)}^2} \\ =2.15\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the current density is$2.15\times {10}^6\mathrm{A}/{\mathrm{m}}^2$ .

(c) Calculation of the magnitude of the electric field

The magnitude of the electric field within the wire can be calculated using the given data in equation (v) as follows:

$$\begin{gathered} \mathrm{E}=\frac{1.20\mathrm{V}}{33.0\mathrm{m}} \\ =3.63\times {10}^{-2}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the value of the field is $3.63\times {10}^{-2}\mathrm{V}/\mathrm{m}$.

(d) Calculation of the rate at which thermal energy will appear in the wire

The rate at which the thermal energy will appear in the wire can be given using equation (vi) as follows:

$$\begin{gathered} \mathrm{P}=1.74\mathrm{A}\times 1.20\mathrm{V} \\ =2.09\mathrm{W} \end{gathered}$$

Hence, the value of the rate of the thermal energy is 2.09 W.