- #1
Jesssa
- 51
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Given a cylindrical wire of radius r, length L, carrying a current I, find the total energy stored inside the wire.
From griffiths,
uem= εE2/2 +B2/2μ
and the tot energy is
∫uem 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/r2 (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=(εE2/2 +B2/2μ)[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/r2 (since dV = x dx dz dthi in cylindrical)
= (2/3) K x3 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)x3 in the second approach since if you put x=r here it will be different to the first approach because of the (2/3)
From griffiths,
uem= εE2/2 +B2/2μ
and the tot energy is
∫uem 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/r2 (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=(εE2/2 +B2/2μ)[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/r2 (since dV = x dx dz dthi in cylindrical)
= (2/3) K x3 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)x3 in the second approach since if you put x=r here it will be different to the first approach because of the (2/3)