- #1
cuongbui1702
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that is my solution:
I=J.A=J.∏r^2
Since r =a => J=0 => I=0 ≠ Io
I think this is wrong but i can't answer why this solution was wrong. Help me Please
Why Does My Current Density Calculation Fail Without Integration?
- Thread starter cuongbui1702
- Start date
In summary: I think that is what you are missing.In summary, the solution to this problem involved using the integral equation, as the current density varies with radius. Integration was necessary to accurately calculate the total current, as it involves summing up small bits of current. This method may have been different from previous exercises, but it was necessary in this case.
that is my solution:
I=J.A=J.∏r^2
Since r =a => J=0 => I=0 ≠ Io
I think this is wrong but i can't answer why this solution was wrong. Help me Please
- #2
berkeman
Mentor
- 68,297
- 22,016
cuongbui1702 said:
that is my solution:
I=J.A=J.∏r^2
Since r =a => J=0 => I=0 ≠ Io
I think this is wrong but i can't answer why this solution was wrong. Help me Please
You need to write and solve the integral equation in order to show that the total current = Io...
- #3
cuongbui1702
- 31
- 0
I read solution, and they also used the integral equation. I do other exercise, i only took I=J.A, and i had a right result. But in this problem, i did not know use the integral equation, why i need to use that?berkeman said:
- #4
berkeman
Mentor
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cuongbui1702 said:
Because the current density varies with radius. When something varies like that, to calculate the total accurately, you need to use integration. Does that make sense?
- #5
tms
- 644
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You need to use the integral because that is what you need to do to solve the problem. You are given the current density as a function of radius, and need to find the total current. When you have a problem like that, you integrate. That's what integration is, a kind of sum, and you here have to sum up a bunch of little bits of current into the total current.cuongbui1702 said:
FAQ: Why Does My Current Density Calculation Fail Without Integration?
What is current density?
Current density is a measure of the amount of electric current flowing through a certain area. It is calculated by dividing the magnitude of the current by the area it is flowing through.
Why is current density important?
Current density is important because it helps us understand how much current is flowing through a specific area, which is crucial for many applications in electronics and engineering. It also allows us to calculate and analyze the distribution of current in a circuit.
What are the units of current density?
The units of current density are amperes per square meter (A/m²) in the SI system or amperes per square centimeter (A/cm²) in the CGS system.
How is current density related to electric field?
Current density is directly proportional to electric field. This means that as the electric field strength increases, the current density also increases. This relationship is described by Ohm's Law: J = σE, where J is current density, σ is the conductivity of the material, and E is the electric field strength.
What factors can affect current density?
The factors that can affect current density include the material's conductivity, the cross-sectional area through which the current is flowing, and the magnitude of the electric field. Temperature and the presence of impurities can also affect current density in some materials.