Solution to Question 5
Problem Statement Analysis:
We have two particles of equal mass $m$ and opposite charges $+q$ and $-q$, initially separated by a distance $r_0$ in a uniform magnetic field $B$ perpendicular to the line joining them. They are released from rest. We need to find the minimum separation between them.
Physical Principles:
The system is subject to the internal Coulomb attraction and the external Magnetic Lorentz force. Since the net external force on the system is zero (the magnetic forces on $+q$ and $-q$ cancel as vectors if $\mathbf{v}_1 = -\mathbf{v}_2$), the center of mass remains stationary. We can analyze the motion in the center of mass frame.
1. Conservation of Energy:
Let $v$ be the speed of each particle at the moment of closest approach. Initially, the particles are at rest. The change in potential energy equals the gain in kinetic energy.
$$ \Delta K + \Delta U = 0 $$ $$ \left( \frac{1}{2}mv^2 + \frac{1}{2}mv^2 \right) – 0 + \left( -\frac{kq^2}{r_{min}} – \left( -\frac{kq^2}{r_0} \right) \right) = 0 $$ $$ mv^2 = \frac{kq^2}{r_{min}} – \frac{kq^2}{r_0} \quad \text{— (1)} $$Where $k = \frac{1}{4\pi\epsilon_0}$.
2. Conservation of Canonical Momentum:
Due to the magnetic field, mechanical momentum is not conserved, but canonical momentum is. Alternatively, we can integrate the equation of motion. The force equation in the y-direction (perpendicular to the line joining them) is:
$$ F_y = qv_x B $$ $$ m \frac{dv_y}{dt} = q \frac{dx}{dt} B $$Integrating with respect to time (from release to closest approach):
$$ m v_y = q B (x – x_0) $$At closest approach, the velocity is purely transverse (perpendicular to the separation vector). If the particles have moved distance $x$ towards the center, the separation is $r_{min} = r_0 – 2x$, so $x = (r_0 – r_{min})/2$. Thus:
$$ v = \frac{qB}{m} \left( \frac{r_0 – r_{min}}{2} \right) \quad \text{— (2)} $$3. Solving for Minimum Separation:
Substitute (2) into (1):
$$ m \left[ \frac{qB}{2m} (r_0 – r_{min}) \right]^2 = \frac{q^2}{4\pi\epsilon_0} \left( \frac{1}{r_{min}} – \frac{1}{r_0} \right) $$ $$ \frac{q^2 B^2}{4m} (r_0 – r_{min})^2 = \frac{q^2}{4\pi\epsilon_0} \frac{r_0 – r_{min}}{r_0 r_{min}} $$Assuming $r_0 \neq r_{min}$, we cancel $(r_0 – r_{min})$:
$$ \frac{B^2}{4m} (r_0 – r_{min}) = \frac{1}{4\pi\epsilon_0 r_0 r_{min}} $$Rearranging to form a quadratic equation for $r_{min}$:
$$ \frac{\pi \epsilon_0 B^2 r_0}{m} (r_0 r_{min} – r_{min}^2) = 1 $$ $$ r_0 r_{min} – r_{min}^2 = \frac{m}{\pi \epsilon_0 B^2 r_0} $$ $$ r_{min}^2 – r_0 r_{min} + \frac{m}{\pi \epsilon_0 B^2 r_0} = 0 $$Solving the quadratic equation $ax^2 + bx + c = 0$ for $r_{min}$:
$$ r_{min} = \frac{r_0 \pm \sqrt{r_0^2 – 4 \left( \frac{m}{\pi \epsilon_0 B^2 r_0} \right)}}{2} $$Since we are looking for the separation that decreases from $r_0$, we consider the physical context. However, the standard form simplifies by factoring out $r_0$. The solution corresponding to the stable approach geometry is:
$$ r_{min} = \frac{r_0}{2} \left[ 1 + \sqrt{ 1 – \frac{4m}{\pi \epsilon_0 r_0^3 B^2} } \right] $$