Field of long cylindrical shell conductor

In summary: The correct value is 3.6E4, not 3.6E3. Therefore, when I plugged it into the equation, I got the correct answer of 5.86E-5 T. In summary, the magnitude of the magnetic field at r=(a+b)/2=5.5 cm (halfway between the two shells) is 5.86E-5 T. This is found by using the formula B=J*(mu_0*r)/2, where J=3.6E4 A/m^2 and r=0.055 m.
  • #1
ttiger2k7
58
0
[SOLVED] Field of long cylindrical shell conductor

Homework Statement


Shown above is the cross-section of a long cylindrical shell conductor of inner radius a=4.00 cm and outer radius b=7.00 cm which carries a current into the page.
The current density J (current/area) is uniform across the shell from r=a to r=b and has the magnitude J=3.6E3 A/m2 where r is the distance from the axis of the shell. (The magnitude of the total current carried by the conductor is thus given by I=J*π*(b2-a2)=37.3 A.)

(a) Find the magnitude of the magnetic field at r=2b=14 cm.

(b) Find the magnitude of the magnetic field at r=(a+b)/2=5.5 cm (halfway between the two shells).

(Hint: how much current is enclosed by the Amperian loop in each case?)


Homework Equations



B = (uIr)/(2pi*R^2) (inside conductor, r<R)
B = (uI)/2pi*r (outside conductor r>R)

where u = 4*pi*10^-7

The Attempt at a Solution



I already solved for part a, getting 5.33E-5 T using the second relevant equation.

But when I used the first relevant equation for part 1, I got 8.4 E-5 which I found out isn't correct. Here's how I did it:

(u*.055*37.3)/(2*pi*(.07^2)) = 8.4E-5 T.

What am I doing wrong?
 

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  • #2
So I don't have a clue what you are plugging in. Most of the people here will be able to follow algebra much easier than seemingly random numbers, just for a future pointer. Anyway, how would you rewrite that first equation in terms of the current density, which is given to you?
 
  • #3
Thanks for your help.

Here's what I got:

Since

[tex]J=\frac{I}{\pi*R^{2}}[/tex]

Then
[tex]B=\frac{\mu_{0}Ir}{2\pi*R^{2}}[/tex]

becomes

[tex]B=J*\frac{\mu_{0}*r}{2}[/tex]

and so

[tex]J=\frac{2B}{\mu_0*r}[/tex]
 
Last edited:
  • #4
Sorry for the bump. I still haven't figured out the problem. Is there anyone else that could help?
 
  • #5
Sorry about leaving you hanging, didn't realize you responded. That should be right.

Inside:
[tex]\mathbf{B}(r) = \frac{\mu_0 J r}{2} \hat{r}[/tex]

Did you get your units right?
 
Last edited:
  • #6
Hm, I'm still not getting the right answer. Just to let you know, the answer I should be getting is 5.86E-5 T.

Using the formula you gave me, I did as followed:

1) For [tex]r[/tex] I used 5.5 cm = .055 m

2) and then I plugged in 3.6E3 for J (since it was given in the problem).

3)[tex]\mu_{0}=4\pi*10^{-7}[/tex]

[tex]\frac{\mu_{0}*3.6*10^{3}*.055}{2}=1.244*10^{-4} T[/tex]

Am I using the wrong numbers?
 
  • #7
I figured it out. I had been using the incorrect value for J.
 

1. What is a field of long cylindrical shell conductor?

A field of long cylindrical shell conductor refers to the electric field that is present around a long, hollow, cylindrical conductor. This type of conductor is commonly used in electrical engineering and physics experiments to study the properties of electric fields.

2. How is the field of long cylindrical shell conductor calculated?

The field of long cylindrical shell conductor is calculated using Gauss's law, which states that the electric flux through a closed surface is directly proportional to the enclosed charge. For a long cylindrical shell conductor, the electric field at a point outside the conductor is equal to the surface charge density divided by 2 times the permittivity of free space.

3. What factors affect the field of long cylindrical shell conductor?

The field of long cylindrical shell conductor is affected by the radius of the conductor, the charge density, and the permittivity of free space. Additionally, the distance from the conductor also plays a role in determining the strength of the electric field.

4. How does the field of long cylindrical shell conductor behave at different points?

The field of long cylindrical shell conductor behaves differently at different points. At points outside the conductor, the electric field is directly proportional to the distance from the center of the conductor. At points inside the conductor, the electric field is zero, as all the charge is on the surface of the conductor.

5. What are some real-world applications of the field of long cylindrical shell conductor?

The field of long cylindrical shell conductor has many real-world applications, including in the design of high-voltage transmission lines, electronic devices, and medical equipment. It is also used in experiments to study the behavior of electric fields and in engineering projects to protect against lightning strikes.

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