Please help and do it step by step all parts of questions and use:
T = time(s)
I=current (amps)
Q=net charge on object (Coulombs)
N=no electrons (and write if added or removed from object and how)
E=elementary charge
Two copper wires of diameter 1 mm and 2 mm are joined end to end in a circuit.
a) Calculate the ratio of the mean drift velocities of the electrons in the two wires if the same current passes through each of them.
b) In which wire will the electrons move faster and why?
As physics - bobpursley Tuesday, December 12, 2017 at 10:25am
you changed the area by 4x.
current=driftvelocity*electrondensitypervolume*area
so if electron density/volume is the same, and area of one is 4x, then drift velociyt is 1/4 because current is constant.
At first before your response i converted to metre so 0.001m and 0.002m and then i found the area for both so for the first it is 7.85 X 10^-7 and the second it is 3.14 X 10^-6. then i did I=Anev for both wires
Copper wire 1:
n= work out
i= SAME CONSTANT
A=7.85 X 10^-7
e= 1.6 X 10^-19
v=??
copper wire 2:
n=work out
i=Same CONSTANT
A=3.14 X 10^-6
e= 1.6 X 10^-19
v=??
and o couldnt work out either as more than one value was missing but i tried to make i equal 1 as their both the same buuit still 2 values missing. Then i read Bobs comment and got confused about where the 4 came from and am still confused with both parts to this question please help!!
give me a break. If you double diameter, area is quadrupled (4x). You are doing a lot of math for no purpose.
you changed the area by 4x.
current=driftvelocity*electrondensitypervolume*area
so if electron density/volume is the same, and area of one is 4x, then drift velociyt is 1/4 because current is constant.
in which wire will the electrons move faster and why?
To solve these questions, we need to use the formula for current (I), drift velocity (v), electron density (n), elementary charge (e), and area (A):
Current (I) = drift velocity (v) * electron density (n) * area (A)
Let's break down the steps to calculate the required values:
a) Calculate the ratio of the mean drift velocities:
1. Calculate the areas of the two copper wires:
- Copper wire 1: diameter of 1 mm (or radius of 0.5 mm = 0.0005 m)
- Area = π * (radius)^2 = 3.14159 * (0.0005)^2 = 7.85 × 10^-7 m^2
- Copper wire 2: diameter of 2 mm (or radius of 1 mm = 0.001 m)
- Area = π * (radius)^2 = 3.14159 * (0.001)^2 = 3.14 × 10^-6 m^2
2. Since the same current passes through both wires, the current (I) is constant and can be treated as the same value for both wires.
3. Use the formula for current to calculate the drift velocity (v):
- Copper wire 1: I = v * n * A1, where n is the electron density per volume
- Copper wire 2: I = v * n * A2
We can rearrange the equation to solve for v:
- Copper wire 1: v1 = I / (n * A1)
- Copper wire 2: v2 = I / (n * A2)
4. Calculate the ratio of the mean drift velocities:
- Ratio = v2 / v1
b) Determine in which wire the electrons will move faster:
The mean drift velocity (v) of electrons is inversely proportional to the cross-sectional area (A) of the wire. So, as the area increases, the drift velocity decreases.
Since the area of Copper wire 2 is 4 times larger than that of Copper wire 1, the drift velocity in Copper wire 2 will be 1/4th of that in Copper wire 1.
Therefore, the electrons will move faster in Copper wire 1 because it has a smaller cross-sectional area.
I hope this step-by-step explanation helps you understand how to solve these questions.