Question

The terminals of a $0.70\mathrm{V}$ watch battery are connected by a $100m$ long gold wire with a diameter of $0.10\mathrm{mm}$. What is the current in the wire?

Short answer

The current in the wire is$2.29mA$.

Step-by-step solution

Given Information

Voltage$=0.70V$

Gold wire length$=100m$

Diameter$=0.10mm$

Explanation 

If we can calculate the potential difference across the wire and the resistance, we can find the current according to Ohm's law.

$I=\frac{U}{R}\to \left(1\right)$

The resistance of the conductor appears in the following expression:

$R=\rho \frac{L}{A}\to \left(2\right)$

Substituting the second expression into the first one,

$I=\frac{AU}{\rho L}.$

The cross-section area is a circle, we can substitute $A=\frac{\pi D^2}{4}$to find

$I=\frac{\pi U{D}^2}{4\rho L}$

Substitute the expression,

role="math" localid="1649055691425" $I=\left(\frac{\pi \times 0.7V\times 0.{0001}^{2}mm}{4\times 2.4\times {10}^{-8}\Omega m\times 100m}\right)$

$I=2.29\mathrm{mA}$

Final Answer 

Hence, the current in the wire is$2.29mA.$