Calculating Current in a Cylindrical Region

AI Thread Summary
The discussion focuses on calculating the current in a cylindrical region with varying current density. For the first case, the current density J decreases linearly from a maximum at the center to zero at the surface, requiring integration to find the total current. In the second scenario, where J is maximum at the surface and decreases to zero at the center, the current calculation yields a different result. Participants emphasize the importance of integrating the contributions from each differential area due to the non-uniform current density. Understanding these differences in current density profiles is crucial for accurate calculations.
Aviegaille
Messages
7
Reaction score
0

Homework Statement



(a)The current density across a cylindrical region of radius R varies according to the equation: J=J0(1-r/R), where r is the distance from the axis of the cylinder. The current density is the maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R. Calculate the current in terms J0 and the region's cross sectional area A=pi*R^2.

(b) Now suppose that a current density was a maximum Jo at the surface and decreased linearly to zero at the axis, so that: J=J0 r/R. Calculate the current. Why is the result different for these two cases?

Homework Equations



I=JA

The Attempt at a Solution



I uploaded a picture of the first part but I am not sure if it's correct. I also don't know how to get the area from this problem. I am thinking of plugging the value of R from I to get the area but I am pretty sure it is not right.
 

Attachments

  • 10711264_757361231009229_384913907_n.jpg
    10711264_757361231009229_384913907_n.jpg
    35.7 KB · Views: 549
Physics news on Phys.org
You cannot just use the total cross-section because the current density is different in different parts of the cylinder.
Instead you have to add up the contributions from each small part of the area.
i.e. you need to set up an integral.

If I is the current and J is the current density, then dI = J.dA
 
Simon Bridge said:
You cannot just use the total cross-section because the current density is different in different parts of the cylinder.
Instead you have to add up the contributions from each small part of the area.
i.e. you need to set up an integral.

If I is the current and J is the current density, then dI = J.dA

Can you elaborate how can I use that equation?
 
You integrate both sides.
You need an expression for J in terms of r and an expression for dA in terms of dr.
Hint: how much current passes through the area between r and r+dr?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top