# Magnetic field of a cylinder inside an AC solenoid

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 $$A(t)=\frac{\mu_0I_0}{\pi a^2}\cos(\omega t).$$