Step 1: Motion on the Incline

Consider the particle sliding down the frictionless slope of inclination $\theta$. The forces acting on the particle are gravity ($mg$), the normal reaction from the slope ($N$), and the Lorentz force ($F_m$).

The velocity $v$ is directed down the slope. Since the magnetic field $\vec{B}$ is directed horizontally into the plane (perpendicular to $\vec{v}$), the Lorentz force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is directed perpendicular to the slope and outwards (away from the surface).

The particle leaves the slope when the normal reaction $N$ becomes zero. Balancing forces perpendicular to the incline at this instant: $$ qvB = mg \cos\theta $$ $$ v = \frac{mg \cos\theta}{qB} \quad \dots(1) $$

Using the work-energy theorem (or kinematics) for the distance $l$ slid down the slope: $$ v^2 – 0 = 2(g \sin\theta)l $$ $$ l = \frac{v^2}{2g \sin\theta} \quad \dots(2) $$

Step 2: Cycloidal Motion Analysis

After leaving the slope, the particle moves under gravity $\vec{g}$ and the magnetic field $\vec{B}$. This combination produces a cycloidal path, which is a superposition of a uniform circular motion and a constant linear drift velocity.

Drift Velocity ($\vec{u}$):
The drift velocity due to the external force $\vec{F}_{ext} = m\vec{g}$ is horizontal: $$ u = \frac{mg}{qB} $$

Circular Motion Component:
The instantaneous velocity $\vec{v}$ at lift-off (from Eq. 1) can be decomposed into the drift velocity and the tangential velocity of the circular motion component ($v_c$). $$ \vec{v} = \vec{u} + \vec{v}_{circ} $$ Taking components along the horizontal ($x$) and vertical ($y$) axes at the moment of lift-off:

• Velocity vector: $\vec{v} = (v \cos\theta)\hat{i} – (v \sin\theta)\hat{j}$
• Drift vector: $\vec{u} = u\hat{i} = \frac{mg}{qB}\hat{i}$

The circular velocity component is: $$ \vec{v}_{circ} = \vec{v} – \vec{u} = \left( v \cos\theta – \frac{mg}{qB} \right)\hat{i} – (v \sin\theta)\hat{j} $$

Substituting $v = \frac{mg \cos\theta}{qB}$: $$ v_{circ, x} = \frac{mg \cos^2\theta}{qB} – \frac{mg}{qB} = \frac{mg}{qB}(\cos^2\theta – 1) = -\frac{mg \sin^2\theta}{qB} $$ $$ v_{circ, y} = -\frac{mg \cos\theta \sin\theta}{qB} $$ The speed of the circular component $v_c$ is: $$ v_c = \sqrt{v_{circ,x}^2 + v_{circ,y}^2} = \frac{mg \sin\theta}{qB} \sqrt{\sin^2\theta + \cos^2\theta} = \frac{mg \sin\theta}{qB} $$

Step 3: Calculating $l$ in terms of $h$

The radius of the circular component of the path is given by $R = \frac{v_c}{\omega}$, where $\omega = \frac{qB}{m}$. $$ R = \frac{ \left( \frac{mg \sin\theta}{qB} \right) }{ \left( \frac{qB}{m} \right) } = \frac{m^2 g \sin\theta}{q^2 B^2} $$

The maximum vertical displacement $h$ in a cycloid (peak-to-peak amplitude) is equal to twice the radius of the generating circle ($2R$): $$ h = 2R \implies h = \frac{2 m^2 g \sin\theta}{q^2 B^2} $$ From this, we can isolate the common constants: $$ \frac{m^2 g}{q^2 B^2} = \frac{h}{2 \sin\theta} \quad \dots(3) $$

Now, substitute Eq. (1) into Eq. (2) to find $l$: $$ l = \frac{ \left( \frac{mg \cos\theta}{qB} \right)^2 }{ 2g \sin\theta } = \frac{m^2 g^2 \cos^2\theta}{q^2 B^2 \cdot 2g \sin\theta} = \left( \frac{m^2 g}{q^2 B^2} \right) \frac{\cos^2\theta}{2 \sin\theta} $$

Substitute the expression from Eq. (3): $$ l = \left( \frac{h}{2 \sin\theta} \right) \frac{\cos^2\theta}{2 \sin\theta} $$ $$ l = \frac{h \cos^2\theta}{4 \sin^2\theta} $$