a. How long must a $0.60mm$diameter aluminum wire be to have a $0.50A$current when connected to the terminals of a $1.5\mathrm{V}$flashlight battery?

b. What is the current if the wire is half this length?

Aluminum wire diameter $=0.60mm$

Current $=0.50A$

Voltage$=1.5V$

The length of the conductor appears in the following expression:

$R=\rho \frac{L}{A}$

$\Rightarrow L=\frac{AR}{\rho }$

Due to the current flowing through the wire and the potential difference across the wire, we can find the resistance by using Ohm's law.

$R=\frac{U}{I}$

Substituting the second expression into the first one, we get

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

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

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

role="math" localid="1649051194334" $L=\left(\frac{\pi \times 0.7V\times 0.{0001}^{2}mm}{4\times 2.4\times {10}^{-8}\left(\Omega m\right)\times 0.5A}\right)$

$=30.3\mathrm{m}$

Hence, the length of the aluminum wire is$30.3m$

Aluminum wire diameter$=0.60\mathrm{mm}$

Current$=0.50\mathrm{A}$

Voltage$=1.5\mathrm{V}$

As the current grows in inverse proportion to the resistance and the resistance grows in proportion to the length, halving the length provides a doubling of the current to $1A$.

Hence, the current if the wire is half this length is$1A.$