Solution
Equation of Motion:
The forces acting on the particle are the Lorentz force and the viscous drag. $\vec{B} = -B\hat{k}$, $\vec{v} = v_x\hat{i} + v_y\hat{j}$.
$$ m\frac{d\vec{v}}{dt} = q(\vec{v} \times \vec{B}) – k\vec{v} $$Resolving into components:
$$ m\dot{v}_x = -qBv_y – kv_x $$ $$ m\dot{v}_y = qBv_x – kv_y $$We need to find the total displacement $X$ and $Y$ when the particle stops ($v_x=v_y=0$). The initial velocity is $\vec{u} = u\hat{i}$.
Integration:
Integrate the equations with respect to time from $t=0$ to $t=\infty$. Let $X = \int v_x dt$ and $Y = \int v_y dt$.
X-component:
$$ m \int_0^\infty \dot{v}_x dt = -qB \int_0^\infty v_y dt – k \int_0^\infty v_x dt $$ $$ m(0 – u) = -qB Y – k X \implies mu = kX + qBY \quad \text{…(1)} $$Y-component:
$$ m \int_0^\infty \dot{v}_y dt = qB \int_0^\infty v_x dt – k \int_0^\infty v_y dt $$ $$ m(0 – 0) = qB X – k Y \implies Y = \frac{qB}{k}X \quad \text{…(2)} $$Solving for Displacement:
Substitute (2) into (1):
$$ mu = kX + qB\left(\frac{qB}{k}X\right) = X \left( \frac{k^2 + q^2B^2}{k} \right) $$ $$ X = \frac{muk}{(qB)^2 + k^2} $$ $$ Y = \frac{qB}{k} X = \frac{muqB}{(qB)^2 + k^2} $$The total displacement vector is:
$$ \vec{r} = \hat{i} \left\{ \frac{kmu}{(qB)^2 + k^2} \right\} + \hat{j} \left\{ \frac{qBmu}{(qB)^2 + k^2} \right\} $$