I have an exercise that I can’t solve. I have an infinite, conducting, non-magnetic cylinder of radius $a$ inside an infinite solenoid of radius $b>a$ with an alternating current flowing through it $I(t)=I_0 \cos(\omega t)$. I need to find the total magnetic field inside the conductor using the quasi-stationary approximation.

I know what the field produced by the solenoid is but I have trouble either applying Ohms Law for the current $J$ or finding the electric field using maxwells equation:

$\nabla \cdot \mathbb{E} =0$ $\nabla \times \mathbb{E}= -\frac{4 \pi}{c} n I_0 \omega \sin(\omega t)$

I tried calculating the curl of the curl of the electric field to habe the laplacian but the PDE is too hard. Any suggestions?


You have to use $$ \nabla\times{\bf B}=\mu_0{\bf j}. $$ Inside the solenoid, the magnetic field due to it is along the conductor and the current. This field is constant and is $B_{0}$. Similarly, ${\bf j}=(\frac{I_0}{\pi a^2}\cos(\omega t),0,0)$. Therefore only component of the rotor that survives is the x-component that yields $$ \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}=\mu_0j_x. $$ The other components give $$ \frac{\partial B_z}{\partial x}-\frac{\partial B_x}{\partial z}=0 $$ $$ \frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}=0. $$ So, we can choose $$ {\bf B}=\left(0,-\frac{A(t)}{2}z,\frac{A(t)}{2}y\right)+(B_0,0,0). $$ Putting this into the equation will give $$ A(t)=\frac{\mu_0I_0}{\pi a^2}\cos(\omega t). $$


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