Water flows through a pipe of radius 4 cm at a rate of 3.5 m/s. Suppose each water molecule is singly ionized (highly unrealistic).
a) What is the electric current density associated with the flow of these ions?
b) What is the electric current in the pipe?
current is Coulombs per second. What did you try to do?
To calculate the electric current density associated with the flow of ions in the pipe, we can use the formula:
J = n * q * v
Where:
J = electric current density (A/m^2)
n = number density of ions (ions/m^3)
q = charge of each ion (Coulombs)
v = flow velocity of the ions (m/s)
a) Electric Current Density:
First, let's find the number density of ions (n). Since each water molecule is singly ionized, the number density of ions is equal to the number density of water molecules.
The number density of water molecules can be calculated using the formula:
n = ρ / M
Where:
n = number density of water molecules (molecules/m^3)
ρ = density of water (kg/m^3)
M = molar mass of water (kg/mol)
The density of water is approximately 1000 kg/m^3, and the molar mass of water is approximately 18 g/mol (or 0.018 kg/mol).
n = 1000 kg/m^3 / 0.018 kg/mol ≈ 5.56 x 10^4 molecules/m^3
Since each water molecule is singly ionized, the number density of ions is the same as the number density of water molecules:
n ≈ 5.56 x 10^4 ions/m^3
Next, we need to determine the charge of each ion (q). In this case, each water molecule is highly unrealistic to be singly ionized. However, assuming each ionization gives one elementary charge (e), the charge of each ion can be calculated as:
q = e
Where:
q = charge of each ion (Coulombs)
e = elementary charge ≈ 1.6 x 10^-19 C
q ≈ 1.6 x 10^-19 C
Finally, we have the flow velocity of the ions (v), which is given as 3.5 m/s.
v = 3.5 m/s
Now, we can calculate the electric current density (J) using the formula mentioned at the beginning:
J = n * q * v
J ≈ (5.56 x 10^4 ions/m^3) * (1.6 x 10^-19 C) * (3.5 m/s)
Calculate J based on the given values.
b) Electric Current:
To calculate the electric current in the pipe, we need to determine the total charge passing through the pipe per unit time.
The electric current (I) is given by:
I = J * A
Where:
I = electric current (Amperes)
J = electric current density (A/m^2)
A = cross-sectional area of the pipe (m^2)
In this case, the cross-sectional area of the pipe can be calculated using the formula:
A = π * r^2
Where:
A = cross-sectional area of the pipe (m^2)
r = radius of the pipe (m)
Given the radius of the pipe is 4 cm, we need to convert it to meters:
r = 4 cm = 0.04 m
Now, we can calculate the cross-sectional area (A):
A = π * (0.04 m)^2
Calculate A using the given formula.
Finally, we can find the electric current (I) by multiplying the electric current density (J) and the cross-sectional area of the pipe (A):
I = J * A
Calculate I by multiplying J and A
Keep in mind that the values obtained here assume a highly unrealistic situation where every water molecule is singly ionized, which is not the case in reality. This calculation is just a theoretical exercise.