Constant magnetic field > Magnetic flux > Problem 9.4.10

Statement

$9.4.10.$ a. Determine the magnetic flux through the surface of a semi-infinite cylinder through which a transverse current with linear density $i$ circulates. The radius of the cylinder $R$.

b. With what force do the halves of a long solenoid with current $I$ attract? Radius of the solenoid $R$, number of turns per unit length of the solenoid $n$.

Solution

a) It is known from the problem $9.3.10$ that deep inside such a solenoid the field is parallel to the axis and is equal to:
$$B_1 = \mu_0 i$$
And at the edge of the component parallel to the axis:
$$B_2 = \frac{\mu_0 i}{2}$$
From the Gauss theorem for a magnetic field, the flow entering the end face inside is equal to the flow exiting through the walls and the outer end, hence:
$$\Phi = (B_1 - B_2)\pi R^2$$
$$\Phi = \frac{1}{2}\mu_0 i\pi R^2$$

b) Savchenko does not specify how the solenoid is cut, lengthwise or across, so if you came for this, Savchenko implies that it is cut across the axis.

Now for the solution, let's allocate an element with dimensions dl by dh on one half of the cylinder (no matter where, the halves are very long), it is affected by a force along the axis equal to:
$$F_{||}=B_{\perp}idldh$$
$$F_{||}=B_{\perp}dSIn$$
$$F_{||}=Ind\Phi$$
where $d\Phi$ is the elementary flow through the side. responsibility.
So the whole force is equal to:
$$F=In\Phi=\frac{1}{2}\mu_0\pi (InR)^2$$

This problem has a shorter solution through the energy density, which is equal to the magnetic pressure:
$$P=\omega=\frac{B^2}{2\mu_0}$$
Let's take the field to be $B=\mu_0 i= \mu_0In$

$$F = PS = \frac{1}{2}\mu_0\pi (InR)^2$$

Answer

$$F = PS = \frac{1}{2}\mu_0\pi (InR)^2$$