Ampere's Law-cylindrical shell

[Krikri]

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

The Attempt at a Solution

Using ampere's law we can easy say that for r<R1 the magnetic field is zero because 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 I_enc . So step by step what i did :


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

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

3)so I_enc= $\frac{2π}{3R1}$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 I_enc but for r=R₂ so I_total= $\frac{2π}{3R1}$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?

 

[rude man]

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

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 I_enc . 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

 

[rude man]

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.

 

[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.

thanks for your reply you really helped me