Solution: Force of Interaction between Two Halves of a Conductor
Problem Analysis:
We consider a long cylindrical conductor of radius $r$ carrying a uniform current density $j$. We need to find the force of interaction per unit length between the two halves of the cylinder divided by a plane containing the axis.
We will calculate the magnetic field inside the conductor, determine the force on a volume element due to this field, and then integrate the component of force perpendicular to the dividing plane over the cross-section of one half.
Step 1: Magnetic Field Inside the Conductor
Using Ampere’s Law at a distance $s$ from the axis ($s < r$):
$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} $$ $$ B(2\pi s) = \mu_0 (j \pi s^2) $$ $$ B = \frac{\mu_0 j s}{2} $$The direction of $\vec{B}$ is tangential to the circle of radius $s$.
Step 2: Force on a Volume Element
Consider a small volume element $dV$ at $(s, \theta)$ with unit length $l=1$. Area element $dA = s ds d\theta$.
The Lorentz force on this element is $d\vec{F} = \vec{j} dV \times \vec{B}$.
Since $\vec{j}$ is along the axis and $\vec{B}$ is tangential, they are perpendicular. The force is directed radially inward (towards the axis).
$$ |d\vec{F}| = j B dV = j \left( \frac{\mu_0 j s}{2} \right) (s ds d\theta) = \frac{\mu_0 j^2 s^2}{2} ds d\theta $$Step 3: Calculating Interaction Force
We want the net force exerted by the bottom half on the top half (or vice versa). This corresponds to the component of the radial force perpendicular to the dividing plane (vertical component $F_y$).
The vertical component of the inward radial force at angle $\theta$ is $dF_y = dF \sin\theta$.
Integrating over the upper semi-circle ($s$ from $0$ to $r$, $\theta$ from $0$ to $\pi$):
$$ F_{net} = \int_{0}^{\pi} \int_{0}^{r} \left( \frac{\mu_0 j^2 s^2}{2} ds d\theta \right) \sin\theta $$ $$ F_{net} = \frac{\mu_0 j^2}{2} \left[ \int_{0}^{r} s^2 ds \right] \left[ \int_{0}^{\pi} \sin\theta d\theta \right] $$Evaluating the integrals:
- $\int_{0}^{r} s^2 ds = \frac{r^3}{3}$
- $\int_{0}^{\pi} \sin\theta d\theta = [-\cos\theta]_0^\pi = -(-1) – (-1) = 2$