Applying amperes law and faradays law to a solenoid and energy density

[jb646]

This isn't really a homework problem, I just need to know how to do a problem similar to this one for the final and I don't want to fail, so I posted it here.

A solenoid with n turns per unit length, length l and radius a (l>>a) has a current I(t)==Io sin(wt) [w is omega], where I and w are constant.
a) using ampere's law calculate the magnetic field inside the solenoid.
b) use faraday's law to calculate the induced electric field around the edge of the solenoid
c) calculate the energy density due to the electric and magnetic field inside the solenoid.

BL=uNI
∇xE= -(dB/dt) <--partial derivative of mag field
U=(εo/2)*E^2+(1/2*uo)*B^2

a) For the first part I know that BL=uNI so I tried just getting B=u*n(Iosin(wt)) but there must be more to it.
b) so if that is true do I just stick that into equation 2 and solve for E
c) and if both of those are right, do I just stick them in equation 3 or is there more I have to do for that one too?

 

[NateD]

For part a) you need to use Ampere's law: B = μo*(N*I/l). So B = μo*(n*Io*sin(ωt)/l). For part b) you can then use Faraday's law: ∇xE = - (dB/dt). Plugging in the equation for B, you get ∇xE = - μo*(n*Io*ω*cos(ωt)/l).For part c) you can then use the equation for energy density: U = (εo/2)*E^2 + (1/2*μo)*B^2. Plugging in the equations for E and B, you get U = (εo/2)*[μo*(n*Io*ω*cos(ωt)/l)]^2 + (1/2*μo)*[μo*(n*Io*sin(ωt)/l)]^2. Simplifying this gives U = (εo*μo*n^2*Io^2*ω^2*cos(ωt)^2 + μo^2*n^2*Io^2*sin(ωt)^2)/(2*l^2).