Why Does My Current Density Calculation Fail Without Integration?

[cuongbui1702]

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

 

[berkeman]

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...

 

[cuongbui1702]

You need to write and solve the integral equation in order to show that the total current = Io...

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]

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?

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?

 

[tms]

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?

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.