Field of long cylindrical shell conductor

[SOLVED] Field of long cylindrical shell conductor

1. 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?)

2. 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

3. 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?
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

Attachments

• 1.5 KB Views: 513

Related Introductory Physics Homework Help News on Phys.org
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?

Here's what I got:

Since

$$J=\frac{I}{\pi*R^{2}}$$

Then
$$B=\frac{\mu_{0}Ir}{2\pi*R^{2}}$$

becomes

$$B=J*\frac{\mu_{0}*r}{2}$$

and so

$$J=\frac{2B}{\mu_0*r}$$

Last edited:
Sorry for the bump. I still haven't figured out the problem. Is there anyone else that could help?

Sorry about leaving you hanging, didn't realize you responded. That should be right.

Inside:
$$\mathbf{B}(r) = \frac{\mu_0 J r}{2} \hat{r}$$

Did you get your units right?

Last edited:
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 $$r$$ 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)$$\mu_{0}=4\pi*10^{-7}$$

$$\frac{\mu_{0}*3.6*10^{3}*.055}{2}=1.244*10^{-4} T$$

Am I using the wrong numbers?

I figured it out. I had been using the incorrect value for J.