Question

Consider a copper wire with a cross-sectional area of 1 mm² (similar to your connecting wires ) and carrying 0.3 A of current, which is about what you get in a circuit with a thick-filament bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Short answer

The strength of the very small electric field required to drive this current through the wire is $5\times {10}^{-3}\text{V/m}$.

Step-by-step solution

Given data

The data can be listed as,

• The cross-sectional area of copper wire is, $A=\text{1}{\text{mm}}^{\text{2}}=1\times {10}^{-6}{\text{m}}^2$.
• Current is, $I=\text{0.3}\text{A}$.

Concept

If a current flows in the circuit, it is influenced by many factors, such as the length of the wire, the cross-sectional area of the wire, and the applied voltage on the wire.

Determination of the required electric field

The wire is made of copper metal only.

The required electric field can be determined using the formula as,

$E=\frac{\rho I}{A}$

Here $\rho$is the resistivity of the copper wire whose value is $16.78\times {10}^{-9}\Omega ·\text{m}$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}E& =& \frac{\left(16.78\times {10}^{-9}\Omega ·\text{m}\right)\left(\text{0.3}\text{A}\right)}{\left(1\times {10}^{-6}{\text{m}}^2\right)}\\ & =& 5.03\times {10}^{-3}\Omega ·{\text{m}}^{-1}·\text{A}\\ & & ~5\times {10}^{-3}\text{V/m}\\ & & \end{array}$

Thus, the strength of the very small electric field required to drive this current through the wire is $5\times {10}^{-3}\text{V/m}$.