Wire with current and field problem

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Homework Help Overview

The problem involves calculating the magnetic field generated by a uniformly distributed electric current in a long, straight wire. The wire has a specified diameter and carries a current of 6.0 A. The discussion includes specific points of interest at various radial distances from the wire's center and explores the relationship between current and magnetic field strength.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of equations related to current density and Ampere's Law. There are attempts to derive the magnetic field at different distances from the wire and to understand the manipulation of equations involved in these calculations. Questions arise about specific steps in the derivation process and the correctness of certain expressions.

Discussion Status

Some participants are seeking clarification on the steps taken in the derivation of the equations, particularly regarding the combination of different equations to arrive at the final expressions. There is an acknowledgment of correct approaches in parts (b) and (c), while others express confusion about specific manipulations in part (a).

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the amount of guidance provided. There is an emphasis on understanding the reasoning behind the equations rather than simply applying them.

Callix
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Homework Statement


An electric current is uniformly distributed throughout a long, straight wire that has a diameter of 0.05m. If the current through the wire is 6.0A, calculate the magnitude of the magnetic field:

a). 0.02m radially away from the wire center
b). 0.05m radially away from the wire center
c). What must the current be to create a magnetic field of magnitude 1.0 T and 0.05m radially away from the wire center.

Homework Equations


In solution

The Attempt at a Solution



I was hoping that someone could check my work and make sure that what I have is correct! Any help would be appreciated.
[/B]
a). This one I wasn't too sure about. I looked up some information but I wasn't sure if it applied here.
I found that:
J= \frac{I}{\pi R^2}I_{enc}=J \pi r^2= \frac{Ir^2}{R^2}\int B\cdot dl=\mu_0 I_{enc}=2 \pi r B2\pi rB= \frac{\mu_0 I}{2 \pi R^2}B= \frac{\mu_0 Ir}{2 \pi R^2}
Is this the correct approach? And if so, could someone explain the steps they took and why? It seems they used two equations and then did some algebra to combine both, but I can't seem to understand what they did in the 4th step to get there...b). \int B\cdot dl=\mu_0 I \rightarrow B=\frac{\mu_0 I}{2 \pi r}
Where I = 6.0 A and r = 0.05m

c). I= \frac{2\pi r B}{\mu_0}
Where B = 1.0 T amd r = 0.05m
 
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Callix said:
could someone explain the steps they took and why? It seems they used two equations and then did some algebra to combine both, but I can't seem to understand what they did in the 4th step to get there...
The third step in (a) is wrong. It should read 2πrB = μ0 Ir2/R2. (The last step is right).
Your (b) and (c) look right.
 
rude man said:
The third step in (a) is wrong. It should read 2πrB = μ0 Ir2/R2. (The last step is right).
Your (b) and (c) look right.

Ah okay, thanks! Would you be able to explain to me how they got to that point, too? For some reason I can't seem to understand how they manipulated the two equations to get the result that they did. It would be greatly appreciated! Thanks again
 
Callix said:
Ah okay, thanks! Would you be able to explain to me how they got to that point, too? For some reason I can't seem to understand how they manipulated the two equations to get the result that they did. It would be greatly appreciated! Thanks again
Which of the 5 steps don't you understand? Seems pretty clearly written up to me.
 
rude man said:
Which of the 5 steps don't you understand? Seems pretty clearly written up to me.

I don't understand how they combined them to form the final equation. I understand the equation for current density and the equation for Ampere's Law. I can't seem to understand what was substituted into the other to get the final equation.
 
If current density is I/πR^2 everywhere across a cross-section and the radius of the cross-section I'm interested in is r, what is the current inside r?
 
rude man said:
If current density is I/πR^2 everywhere across a cross-section and the radius of the cross-section I'm interested in is r, what is the current inside r?

I'm not sure... This is why I was asking if you could explain the steps to me
 

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