Question

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current in the wire? Express your answer using three significant figures.

Answer 1

`I=0.047A`

Step-by-step explanation:

Let's use Ohm's law:

`V=IR`

or

`I=(V)/(R)` (1)

Where:

`V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance`

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

`R=\rho*(l)/(A)` (2)

Where:

`R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^(-7) \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^(-8)`

Keep in mind that the electrical resistivity of the gold is a known constant which is
`\rho_g_o_l_d=2.35*10^(-8)` and the cross sectional area of the conductor is calculated as:

`A=\pi *(r^(2))=\pi  *(0.0002m)^(2) =1.256637061*10^(-7) m^(2)`

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

`R=(2.35*10^(-8))*(80)/(1.256637061*10^(-7) )  =14.96056465\Omega`

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

`I=(0.7)/(14.96056465)\approx0.047A`