- #1

- 51

- 0

From griffiths,

u

_{em}= εE

^{2}/2 +B

^{2}/2μ

and the tot energy is

∫u

_{em}dV

I have my E and B fields, but my B field is a function of x where x<r, (E is uniform)

B=kx/r

^{2}(k=all the constants)

my question is,

it says

__inside__the wire, does this mean i cannot put x=r and integrate easily to get

Energy=(εE

^{2}/2 +B

^{2}/2μ)[itex]\pi r^2 L[/itex] ?

Where B = k/r = K

or do i have to integrate B seperately to get something like

∫∫∫Kx x dx dz d[itex]\phi[/itex] where K = k/r

^{2}(since dV = x dx dz dthi in cylindrical)

= (2/3) K x

^{3}L pi

If so would this be it? No bounds on x, giving the energy at some radius inside the wire?

i guess the question is over

__what__volume is considered inside the cylinder, some x<a or just x=a?

Im unsure about this because of the (2/3)x

^{3}in the second approach since if you put x=r here it will be different to the first approach because of the (2/3)