# Ampere's Law-cylindrical shell

#### Krikri

1. Homework Statement
A conducting cylindrical shell has it's axle on Ox axis, inner radius R1 and outer radius R2. If the volume current density is given by the formula $\vec{j}$(r)=J0$\frac{r}{R1}$$\hat{i}$ where J0 is a constant and r the distance from the cylinder's axle, calculate the magnetic field $\vec{B}$ : 1) Everywhere in space and 2) On the Oy. Oz axis in a position y>R2 and z>R2

3. The Attempt at a Solution

Using ampere's law we can easy say that for r<R1 the magnetic field is zero beacause there is no current passing through the amperian loop. For R1<r<R2 the region that the not uniform density takes place.
I draw an amperian loop of radius r. The whole point as i see it is to calculate the Ienc . So step by step what i did :

1) dI=J*da ,da=2πrdr so dI= $\frac{2π}{R1}$j0 r2dr

2)so Ienc= $\frac{2π}{R1}$j0 ∫r2dr from R1 to r.

3)so Ienc= $\frac{2π}{3R1}$j0 (r3-R13)

4) Thus the amperes law gives ∫Bdl=μ0$\frac{2π}{3R1}$j0 (r3-R13) and B2πr=μ0$\frac{2π}{3R1}$j0 (r3-R13) so B=μ0$\frac{2π}{3R1}$j0 [r2-(R13/r)]

5) For r>R2, outside the cylinder, the amperian loop will include the total current which the same as Ienc but for r=R2 so Itotal= $\frac{2π}{3R1}$j0 (R23-R13)

6) And applying the ampere's law again into a loop radius r>R2 it gives B= (μ0j0/3r)[(R23/R1)-R12]

7) It seems that in the Oy,Oz axis the magnetic field is the same and outside the cylinder it decreases with a ratio 1/r. The magnetic field is doing loops around the cylinder and inside the area with the current in the $\hat{φ}$ direction.

I am not sure if what i did is correct and the conclusions i made. Can you comment please?

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#### rude man

Homework Helper
Gold Member
1. Homework Statement
A conducting cylindrical shell has it's axle on Ox axis, inner radius R1 and outer radius R2. If the volume current density is given by the formula $\vec{j}$(r)=J0$\frac{r}{R1}$$\hat{i}$ where J0 is a constant and r the distance from the cylinder's axle, calculate the magnetic field $\vec{B}$ : 1) Everywhere in space and 2) On the Oy. Oz axis in a position y>R2 and z>R2

3. The Attempt at a Solution

For R1<r<R2 the region that the not uniform density takes place.
I draw an amperian loop of radius r. The whole point as i see it is to calculate the Ienc . So step by step what i did :

1) dI=J*da ,da=2πrdr
Right.
so dI= $\frac{2π}{R1}$j0 r2dr
Wrong. Why did you not integrate dI = 2 pi j r' dr' from R1 to r?

#### Krikri

Right.

Wrong. Why did you not integrate dI = 2 pi j r' dr' from R1 to r?
I thought that if the current is in the green area, then the integration is from R1 o r. EDIT: i saw it now. I did integrate from R1 to r..i forgot to put the 'd' in the r2..so it is dr2 in the first step

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#### rude man

Homework Helper
Gold Member
I did integrate from R1 to r..i forgot to put the 'd' in the r2..so it is dr2 in the first step
Not sure I understand that.

Actually however, you did everything right in your original post except when it came to evaluating B for case 2 where you forgot to divide by 2 pi.

I'm sorry, your expressions fooled me for a while but they were mostly OK all along.

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#### Krikri

Not sure I understand that.
never mind that, it was a mistake i thought i did :)

Actually however, you did everything right in your original post except when it came to evaluating B for case 2 where you forgot to divide by 2 pi.

I'm sorry, your expressions fooled me for a while but they were mostly OK all along.