Constant magnetic field > The magnetic field from a current distributed over a surface of space > Problem 9.3.21

Statement

$9.3.21.$ Determine the magnetic field induction in a long cylindrical cavity located inside a cylindrical conductor,
if the axis of the cavity is parallel to the axis of the conductor and is located at a distance $d$ from it. The
current is distributed evenly over the cross-section of the conductor. Current density $j$.

Solution

In previous problems, we derived a formula for the induction at a distance from the axis inside a cylindrical wire:
$$B = \frac{\mu_0 j x}{2}$$
However, we did not mention one remarkable property of this formula: since the induction is directed tangentially to the circular contours along the axis, mathematically this can be rewritten in vector form:
$$\hat{B} = \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$
Let's use an equally remarkable property of the vector product:
$$(\hat{a}+\hat{b})\times \hat{c} = \hat{a}\times \hat{c} + \hat{b} \times \hat{c}$$
Let's take a point at a distance x from the cavity axis; in vector form from the wire axis, it is located on the vector $\hat{r}=\hat{d}+\hat{x}$. We represent the field from the wire with the cavity as the difference between the fields of the small and large cylinders in vector form. Taking into account the properties of the vector product, we obtain:
$$\hat{B} = \hat{B} = \frac{\mu_0[\hat{j}\times \hat{d}]}{2}+\frac{\mu_0[\hat{j}\times \hat{x}]}{2} - \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$
$$B = \frac{\mu_0 j d}{2}$$

Answer

$$B = \frac{\mu_0 j d}{2}$$