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## 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 [itex]\vec{j}[/itex](r)=J

_{0}[itex]\frac{r}{R1}[/itex][itex]\hat{i}[/itex] where J

_{0}is a constant and r the distance from the cylinder's axle, calculate the magnetic field [itex]\vec{B}[/itex] : 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 beacause there is no current passing through the amperian loop. For R

_{1}<r<R

_{2}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= [itex]\frac{2π}{R1}[/itex]j

_{0}r

^{2}dr

2)so I

_{enc}= [itex]\frac{2π}{R1}[/itex]j

_{0}∫r

^{2}dr from R

_{1}to r.

3)so I

_{enc}= [itex]\frac{2π}{3R1}[/itex]j

_{0}(r

^{3}-R

_{1}

^{3})

4) Thus the amperes law gives ∫Bdl=μ

_{0}[itex]\frac{2π}{3R1}[/itex]j

_{0}(r

^{3}-R

_{1}

^{3}) and B2πr=μ

_{0}[itex]\frac{2π}{3R1}[/itex]j

_{0}(r

^{3}-R

_{1}

^{3}) so B=μ

_{0}[itex]\frac{2π}{3R1}[/itex]j

_{0}[r

^{2}-(R

_{1}

^{3}/r)]

5) For r>R

_{2}, outside the cylinder, the amperian loop will include the total current which the same as I

_{enc}but for r=R

_{2}so I

_{total}= [itex]\frac{2π}{3R1}[/itex]j

_{0}(R

_{2}

^{3}-R

_{1}

^{3})

6) And applying the ampere's law again into a loop radius r>R2 it gives B= (μ

_{0}j

_{0}/3r)[(R

_{2}

^{3}/R

_{1})-R

_{1}

^{2}]

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 [itex]\hat{φ}[/itex] direction.

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