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Induced Current Density On Disk Due to Changing Magnetic Field

  • Thread starter darkfall13
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  • #1
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Homework Statement



A very thin conducting disc of radius [tex]a[/tex] and conductivity [tex]\sigma[/tex] lies in the x-y plane with the origin at its center. A spatially uniform induction is present and given by [tex]B=B_0 cos\(\omega t \)\hat{z}[/tex]. Find the induced current density [tex]\vec{J}[/tex] in the disc.

Homework Equations



[tex] \vec{J} = \sigma ( \vec{E} + \vec{v} \times \vec{B} ) [/tex]
[tex] \vec{\nabla} \times \vec{B} = \mu_0 \vec{J} [/tex]
[tex] \vec{\nabla} \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}[/tex]

The Attempt at a Solution



I'm utterly confused on this problem, I really just don't know where to start. The first equation for [tex]\vec{J}[/tex] is utterly useless as our medium is stationary. The second is also not too helpful as it doesn't take into account the magnetic field's time dependence. The third also suffers from this problem. So that is my first hiccup. I know it's zero progress but this subject is hard for me :(
 

Answers and Replies

  • #2
33
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I'm silly, I think I got it started:

[tex] \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{\partial t} = + B_0 \omega \sin(\omega t) \hat{z} [/tex]
[tex] \oint_C \vec{E} \cdot d\vec{s} = \oint_C E_\phi \hat{\phi} \cdot \rho d\phi \hat{\phi} = 2 \pi \rho E_\phi [/tex]
[tex] = \int_S (\vec{\nabla} \times \vec{E}) \cdot d\vec{a} = \omega B_0 \sin(\omega t) \int da_z [/tex]
[tex] E_\phi = \frac{1}{2} \omega B_0 \sin(\omega t) [/tex]
 

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