9.3.20. In an infinite plate of thickness $h$, a cylindrical cavity of radius $h/2$ was cut out, the axis of which is parallel to the surfaces of the plate. In the entire volume of the plate, with the exception of the cavity, a current flows along the axis of the cavity. Find the distribution of magnetic field induction along a straight line $OA$ that passes through the cavity axis and is perpendicular to the plate surfaces. Current density $j$.
Solution
In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively: $$B^{in}_{full} = \mu_0 j x$$ $$B^{out}_{full}=\frac{\mu_0 j h}{2}$$ $$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$ $$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$
Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers:
For $0<x<h/2$: $$B = \frac{\mu_0 j x}{2}$$
For $x > h/2$: $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$
Answer
For $0<x<h/2$: $$B = \frac{\mu_0 j x}{2}$$
For $x > h/2$: $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$
