How do I solve this problem:
A particular wire has a current of 10A.
A) How many electrons pass a fixed point every 5 seconds?
B)If a wire has a radius of 2 x 10^(-3) m and the electron drift speed is 2.63 x 10^(-5) m/s, determine the number density of electrons in the wire.
The teacher posts the answers for us, so the answer for
A) 3.125 x 10^(20) e^(-)
B) 1.89 x 10^(29) e^(-)/ m^3
It looks like you are switching names. Why?
A) Current (in Amperes) is charge passing by per second.
B) Current = (free electron number density)x (drift speed) x (cross sectional area)
Solve for the FREE electron density. (Electrons that remain attached to the nuclei of the conductor atoms are not included).
Thanks for the help! And BTW, I am not switching my name.
To solve this problem, we can use the formula for current, which is given by:
I = n * A * v * q
Where:
I = current (in Amperes)
n = number density of electrons (in electrons per cubic meter)
A = cross-sectional area of the wire (in square meters)
v = drift speed of the electrons (in meters per second)
q = charge of one electron (1.6 x 10^-19 C)
Let's solve this problem step by step:
A) How many electrons pass a fixed point every 5 seconds?
To find the number of electrons passing a fixed point, we need to find the total charge passing through that point. The charge passing through a point is given by the product of current and time:
Q = I * t
Where:
Q = charge (in coulombs)
I = current (in Amperes)
t = time (in seconds)
Since the charge of one electron is given by q = 1.6 x 10^-19 C, the number of electrons passing the fixed point can be calculated by dividing the total charge by the charge of one electron:
Number of electrons = Q / q
Given that the current is 10A and the time is 5 seconds, we can substitute the values into the equations:
I = 10A
t = 5s
q = 1.6 x 10^-19 C
1. Calculate the charge passing through the fixed point:
Q = I * t
Q = 10A * 5s = 50C
2. Calculate the number of electrons passing the fixed point:
Number of electrons = Q / q
Number of electrons = 50C / (1.6 x 10^-19 C) = 3.125 x 10^20 electrons
Therefore, the number of electrons passing the fixed point every 5 seconds is 3.125 x 10^20 electrons.
B) Determine the number density of electrons in the wire.
To find the number density of electrons in the wire, we can rearrange the formula for current:
n = I / (A * v * q)
Given that the current is 10A, the radius of the wire is 2 x 10^(-3) m, and the drift speed of electrons is 2.63 x 10^(-5) m/s, we can substitute the values into the equation:
I = 10A
A = π * r^2 (where r is the radius of the wire)
v = 2.63 x 10^(-5) m/s
q = 1.6 x 10^(-19) C
1. Calculate the area of the wire:
A = π * r^2
A = π * (2 x 10^(-3))^2
A ≈ 3.14 * 4 x 10^(-6) m^2
A ≈ 1.256 x 10^(-5) m^2
2. Calculate the number density of electrons:
n = I / (A * v * q)
n = 10A / (1.256 x 10^(-5) m^2 * 2.63 x 10^(-5) m/s * 1.6 x 10^(-19) C)
n ≈ 1.89 x 10^29 electrons/m^3
Therefore, the number density of electrons in the wire is approximately 1.89 x 10^29 electrons/m^3.
To solve these problems, you need to understand the basic concepts and formulas related to electric current, charge, and density of electrons.
A) How many electrons pass a fixed point every 5 seconds?
To calculate the number of electrons passing a fixed point, you need to use the formula that relates current (I) and charge (Q).
The formula is Q = I * t, where Q is the charge in Coulombs, I is the current in Amperes, and t is the time in seconds.
In this case, you know the current = 10A and the time = 5s. So, you can substitute these values into the formula:
Q = 10A * 5s = 50C
Every 1 Coulomb (C) contains approximately 6.24 x 10^18 electrons. Therefore, to find the number of electrons passing the fixed point, you can multiply the charge (Q) by this conversion factor:
Number of electrons = Q * (6.24 x 10^18) = 50C * (6.24 x 10^18) ≈ 3.125 x 10^20 electrons
Therefore, the answer to part A is approximately 3.125 x 10^20 electrons.
B) Determine the number density of electrons in the wire.
To find the number density of electrons in the wire, you need to use the formula that relates current (I), drift velocity (v_d), cross-sectional area (A), electronic charge (e), and number density (n).
The formula is n = I / (e * A * v_d), where n is the number density of electrons in m^-3, I is the current in Amperes, e is the elementary charge (1.6 x 10^(-19) C), A is the cross-sectional area in m^2, and v_d is the electron drift velocity in m/s.
In this case, you know the current (I) = 10A, the radius (r) = 2 x 10^(-3)m, and the electron drift speed (v_d) = 2.63 x 10^(-5)m/s.
To calculate the cross-sectional area (A) of the wire, you need to use the formula for the area of a circle:
A = π * r^2
A = π * (2 x 10^(-3))^2 ≈ 12.57 x 10^(-6)m^2
Now you can substitute the known values into the formula for number density (n):
n = 10A / (1.6 x 10^(-19)C * 12.57 x 10^(-6)m^2 * 2.63 x 10^(-5)m/s)
n ≈ 1.89 x 10^29 m^-3
Therefore, the answer to part B is approximately 1.89 x 10^29 electrons per cubic meter (e^-/m^3).