- #1
richyw
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Homework Statement
A long solenoid with 1000 turns per meter and radius 2cm carries an oscillating current given by [itex](5A)\sin(100\pi t)[/itex]. What is the electric field induced at radius r=1cm from the axis of the solenoid? What is the direction of this electric field when the current is increasing counterclockwise in the coil?
Homework Equations
[tex]\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_B}{dt}[/tex]
[tex]\Phi_B=\oint\vec{B}\vec{dA}[/tex]and I have this on my formula sheet so I will start at this point (magnetic field inside a solenoid with n turns per unit length and current I[tex]B=\mu_0nI[/tex]
The Attempt at a Solution
So I think in this situation I can say [tex]\Phi_B=BA=(\mu_0nI)(\pi r^2)[/tex]So[tex]\frac{d\Phi_B}{dt}=\pi\mu_0nr^2\frac{dI}{dt}[/tex]
Also
[tex]\oint\vec{E}\cdot d\vec{l}=2E\pi r[/tex]
So I got (ignoring the negative, just considering the magnitude)
[tex]E=\frac{\pi\mu_0nr^2}{2\pi r}\frac{dI}{dt}= \frac{1}{2} \mu_0nr\frac{dI}{dt}=\frac{1}{2}\mu_0nr\omega I_{max}\cos(\omega t)[/tex]where [itex]\omega=100\pi[/itex] and [itex]I_{max}=5A[/itex]
This makes sense to me but is not getting me the correct answer! Also I only reasoned that the electric field would be clockwise, but not really sure on that part. Does anyone see where I am going wrong?