Constant magnetic field > The magnetic field from a current distributed over a surface of space > Problem 9.3.22

Statement

$9.3.22.$ a. Two cylinders of radius $R$, whose axes are at a distance $a$ from each other, intersect, as shown in the
figure. Currents with a density of $±j$ flow through the shaded areas along the axes in opposite directions.
Find the magnetic field induction in the region lying between the shaded areas.

b. Using the result of the previous problem and applying the limit transition method, find for $a → 0$,
$j → ∞$ the distribution of the linear current density on the surface of a cylinder of radius $R$, which gives
a homogeneous magnetic field of induction $B_0$ inside the cylinder. How is the maximum linear current
density related to the field induction $B_0$?

Solution

a) As in the previous problem, we will use the property of the vector product and the formula for the field inside a cylindrical wire. Consider a point at distances $r_1$ and $r_2$ from the centers of the cylinders:
$$\hat{B} = \frac{\mu_0 [\hat{r}_1 \times \hat{j}]}{2} - \frac{\mu_0 [\hat{r}_2 \times \hat{j}]}{2} = \frac{\mu_0 [\hat{a} \times \hat{j}]}{2}$$
$$B = \frac{\mu_0 a j}{2}$$
b) The detailed solution to this point repeats in all its main points the solution to problem 6.2.14, which I recommend you read. However, it should be mentioned that here it is more convenient to express the answer in terms of a sine, since the induction lines inside are perpendicular to the vector $\hat{a}$.
$$i = j a sin \varphi$$
$$j = \frac{2B_0}{\mu_0 a}$$
$$i = \frac{2B_0}{\mu_0}sin\varphi$$
$$i_{max} = \frac{2B_0}{\mu_0}$$

Answer

а)$$B = \frac{\mu_0 a j}{2}$$
b)$$i = \frac{2B_0}{\mu_0}sin\varphi$$
$$i_{max} = \frac{2B_0}{\mu_0}$$