Calculating Current in a Cylindrical Region

Click For Summary
SUMMARY

The discussion focuses on calculating the current in a cylindrical region with varying current density. For part (a), the current density is defined as J=J0(1-r/R), leading to the integral I=∫J dA, where A=πR². For part (b), the current density is J=J0(r/R), resulting in a different current calculation due to the linear variation of current density. The key takeaway is that the method of integrating the contributions from each differential area is crucial for accurate current calculation.

PREREQUISITES
  • Understanding of current density and its mathematical representation
  • Familiarity with integral calculus for area calculations
  • Knowledge of cylindrical coordinates and their applications
  • Basic principles of electromagnetism, specifically Ohm's Law
NEXT STEPS
  • Study the derivation of current density equations in cylindrical coordinates
  • Learn about the application of integrals in calculating physical quantities
  • Explore variations in current density and their effects on total current
  • Investigate the implications of different current density profiles in practical scenarios
USEFUL FOR

Students and professionals in physics and electrical engineering, particularly those focusing on electromagnetism and circuit analysis, will benefit from this discussion.

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: 558
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
526
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
825
  • · Replies 8 ·
Replies
8
Views
2K