Calculating Current Density & Current in Gold Wire

[kyang002]

The electron drift speed in a gold wire is 3.0 * 10^-4.

1. What is the current density in the wire?

2. What is the current if the wire diameter is 0.50 mm?

I am unsure of which equation to use.
Q = I delta t ??

Can someone help me out?

 

[Nylex]

Is it too hard to post all your questions in one thread?

Also, how on Earth does Q = I delta t help you? How about I = nAqv?

 

[wais]

To calculate the current density in the wire, we can use the equation J = I/A, where J is the current density, I is the current, and A is the cross-sectional area of the wire. Since we are given the electron drift speed (v) in the wire, we can use the equation I = nqvA, where n is the number of free electrons per unit volume, q is the charge of an electron, and A is the cross-sectional area of the wire.

1. Using the given electron drift speed of 3.0 * 10^-4, we can calculate the current density as follows:

J = I/A = (nqvA)/A = nqv

Since we do not have information about the number of free electrons per unit volume, we cannot calculate the exact value of current density. However, we can say that the current density in the gold wire will be directly proportional to the electron drift speed.

2. To calculate the current in the wire, we can use the equation I = nqvA, where n is the number of free electrons per unit volume, q is the charge of an electron, v is the electron drift speed, and A is the cross-sectional area of the wire.

Since we are given the wire diameter, we can calculate the cross-sectional area using the formula A = πr^2, where r is the radius of the wire.

r = 0.50 mm/2 = 0.25 mm = 0.25 * 10^-3 m

A = π(0.25 * 10^-3)^2 = 1.96 * 10^-7 m^2

Now, we need to find the number of free electrons per unit volume (n) in gold. This value can be found in a table of material properties or can be calculated using the density of gold (19.3 g/cm^3) and its atomic mass (196.97 g/mol).

n = (density * Avogadro's number)/atomic mass = (19.3 g/cm^3 * 6.022 * 10^23 mol^-1)/196.97 g/mol = 5.92 * 10^28 electrons/m^3

Substituting the values in the equation I = nqvA, we get:

I = (5.92 * 10^28 electrons/m^3) * (1.6 *