Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.0 A. The resistance per unit length is to be$\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{}\mathbf{\Omega }\mathbf{/}\mathbf{km}$. The densities of copper and aluminum are 8960and$\mathbf{2600}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$, respectively. Compute (a) the magnitude Jof the current density and (b) the mass per unit length$\mathit{\lambda }$for a copper cable and (c) Jfor an aluminum cable (d)for an aluminum cable.

• Current,$i=60.0\mathrm{A}$
• Resistance per unit length,$\frac{R}{L}=0.150\mathrm{\Omega }/\mathrm{km}\mathrm{or}0.150\times {10}^{-3}\mathrm{\Omega }/\mathrm{m}$
• Density of copper,$d=8960\mathrm{kg}/{\mathrm{m}}^2$
• Density of aluminum,$d\text{'}=2600\mathrm{kg}/{\mathrm{m}}^3$

The resistance is directly proportional to the length and inversely proportional to the area of the cross-section of the conductor. The constant of proportionality is called as the resistivity of that material. The current density is the current across the unit area at a given point in the conductor.

First, we have to write the area in terms of resistivity and resistance per unit length. Using this expression of the area in the formula for current density, we can find the magnitude of current density for the copper and aluminum cables. Using the expression for the area and substituting all the corresponding values, we can find the mass per unit length for the copper and aluminum cables.

Formulae:

The resistance of a wire due to its resistivity, $\mathrm{\rho }=\frac{RA}{L}$ …(i)

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

The current density of current passing through the area, $J=\frac{i}{A}$ …(ii)

Here, J is the current density,i is current, and A is the area of cross-section.

The linear density i.e., the mass per unit length,

$\lambda =\frac{m}{L}$ …(iii)

Here,$\lambda$is the linear density,m is the charge and L is the length of the wire.

The density of a substance, $d=\frac{m}{V}$ …(iv)

Here,d is the density,m is the mass and V is the volume.

From equation (i), we can get the value of the cross-sectional area as follows:

$A=\frac{\mathrm{\rho }}{R/L}$ …(v)

The magnitude of the current density for a copper cable is given by using the given data and the above value in equation (ii) as follows. (Since, the resistivity of copper is)

$$\begin{gathered} \mathrm{\rho }=1.69\times {10}^{-8}\mathrm{\Omega m}) \\ \mathrm{J}=\frac{\mathrm{i}}{\left({\displaystyle \frac{\mathrm{\rho }}{\mathrm{R}/\mathrm{L}}}\right)} \end{gathered}$$ …(vi)

Substitute the given values in the equation (vi)

$$\begin{gathered} J=\frac{60.0A}{\left({\displaystyle \frac{1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}{0.150\times {10}^{-3}\mathrm{\Omega }/\mathrm{m}}}\right)} \\ =5.32\times {10}^5\frac{\mathrm{A}}{{\mathrm{m}}^2} \end{gathered}$$

Hence, the magnitude J of the current density for a copper cable is $5.32\times {10}^5\frac{\mathrm{A}}{{\mathrm{m}}^2}$.

The mass per unit length$\lambda$for a copper cable is given using mass value of equation (iv) in equation (iii) as follows:

$$\begin{gathered} \lambda =\frac{Vd}{L} \\ =Ad\left(\because Area=Volume/Lenght\right) \end{gathered}$$

Therefore, we can write,

$\lambda =\frac{\mathrm{\rho }d}{{\displaystyle \frac{R}{L}}}$ …(vii)

Substitute the values to get the

$$\begin{gathered} \lambda =\frac{1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}{0.150\times {10}^{-3}\mathrm{\Omega }/\mathrm{m}}\times 8960\mathrm{kg}/{\mathrm{m}}^3 \\ =100.95\times {10}^{-2}\mathrm{kg}/\mathrm{m} \\ \approx 1.01\mathrm{kg}/\mathrm{m} \end{gathered}$$

Thus, the mass per unit length$\lambda$ for a copper cable is 1.01 kg/m.

The current density for an aluminium cable is given by using the given data in equation (b) as follows: (Since, the resistivity of aluminium is$\mathrm{\rho }=2.75\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}$)

$$\begin{gathered} J=\frac{60.0\mathrm{A}}{\left({\displaystyle \frac{2.75\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}{0.150\times {10}^{-3}\mathrm{\Omega }/\mathrm{m}}}\right)} \\ =3.27\times {10}^5\frac{\mathrm{A}}{{\mathrm{m}}^2} \end{gathered}$$

Hence, the magnitude J of the current density for an aluminium cable is $3.27\times {10}^5\frac{\mathrm{A}}{{\mathrm{m}}^2}$.

The mass per unit length$\lambda$for an aluminum cable is given by substituting the given values in the equation (vii) from part (b) calculations as follows:

$$\begin{gathered} \mathrm{\lambda }=\frac{2.75\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}{0.150\times {10}^{-3}{\displaystyle \frac{\mathrm{\Omega }}{\mathrm{m}}}}\times 2600\frac{\mathrm{kg}}{{\mathrm{m}}^3} \\ =0.4766\mathrm{kg}/\mathrm{m} \\ \approx 0.48\mathrm{kg}/\mathrm{m} \end{gathered}$$

Thus, the mass per unit length for an aluminum cable is 0.48 kg/m.