This problem involves understanding the motion of a charged particle in the presence of both electric and magnetic fields. The total force acting on the particle is given by the Lorentz force, which is the sum of the electric force and the magnetic force.

The electric field is constant and points in the positive z-direction. Let's represent it as \(\vec{E} = E_0 \hat{k}\), where \(E_0\) is a positive constant and \(\hat{k}\) is the unit vector in the positive z-direction.

The magnetic field is constant and points in the positive x-direction. Let's represent it as \(\vec{B} = B_0 \hat{i}\), where \(B_0\) is a positive constant and \(\hat{i}\) is the unit vector in the positive x-direction.

The particle is positively charged, let's say with charge \(q > 0\). Its initial velocity is zero, \(\vec{v}(0) = 0\). The forces acting on the particle are:

• Electric force: \(\vec{F}_E = q\vec{E}\)
• Magnetic force: \(\vec{F}_B = q(\vec{v} \times \vec{B})\)

The total force is \(\vec{F} = \vec{F}_E + \vec{F}_B = q\vec{E} + q(\vec{v} \times \vec{B})\).

Initial Motion of the Particle

Initially, when the particle is at rest (\(\vec{v} = 0\)), the magnetic force is zero (\(\vec{F}_B = 0\)). The only force is the electric force:

\(\vec{F}(0) = q\vec{E} = q(E_0 \hat{k}) = qE_0 \hat{k}\)

Since the particle is positive (\(q>0\)), the initial force is in the positive z-direction. This force will accelerate the particle, causing it to gain velocity in the positive z-direction.

Effect of Magnetic Field and Drift Motion

As the particle gains velocity, say \(\vec{v}\), the magnetic force \(q(\vec{v} \times \vec{B})\) becomes non-zero. The motion of a charged particle in combined perpendicular electric and magnetic fields is generally a superposition of a circular motion and a drift motion.

For constant, uniform, and perpendicular electric and magnetic fields, there is a characteristic drift velocity, \(\vec{v}_d\). This drift velocity is the average velocity of the particle over long periods, representing the overall direction and speed of motion. The formula for the drift velocity is given by:

\(\vec{v}_d = \frac{\vec{E} \times \vec{B}}{B^2}\)

Let's calculate the drift velocity using the given fields:

\(\vec{E} = E_0 \hat{k}\)

\(\vec{B} = B_0 \hat{i}\)

The cross product \(\vec{E} \times \vec{B}\) is:

\(\vec{E} \times \vec{B} = (E_0 \hat{k}) \times (B_0 \hat{i}) = E_0 B_0 (\hat{k} \times \hat{i})\)

Recall the cross products of unit vectors: \(\hat{i} \times \hat{j} = \hat{k}\), \(\hat{j} \times \hat{k} = \hat{i}\), \(\hat{k} \times \hat{i} = \hat{j}\).

So, \(\hat{k} \times \hat{i} = \hat{j}\).

Therefore, \(\vec{E} \times \vec{B} = E_0 B_0 \hat{j}\).

The magnitude of the magnetic field is \(B = |\vec{B}| = \sqrt{B_0^2} = B_0\), so \(B^2 = B_0^2\).

Now, substitute these into the drift velocity formula:

\(\vec{v}_d = \frac{E_0 B_0 \hat{j}}{B_0^2} = \frac{E_0}{B_0} \hat{j}\)

Overall Motion Direction

The drift velocity vector \(\vec{v}_d\) is in the direction of \(\hat{j}\), which is the positive y-direction. The motion of the particle is a cycloidal path that is centered around this drift velocity. At large times, the particle's position will have a net displacement primarily in the direction of this drift.

Thus, the overall motion of the particle at large times is adrift along the positive y-direction.