To find the magnitude and direction of the particle's velocity in the x-y plane, we can use the Lorentz force equation, which describes the net force acting on a charged particle in electric and magnetic fields:

$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$

Where:

• $\mathbf{F}$ is the net force.
• $q$ is the charge of the particle.
• $\mathbf{E}$ is the electric field.
• $\mathbf{v}$ is the velocity of the particle.
• $\mathbf{B}$ is the magnetic field.

Given:

• $\mathbf{E} = 1450 \, \text{N/C}$ (in the positive x-direction)
• $\mathbf{B} = 1.22 \, \text{T}$ (in the positive z-direction)
• $\mathbf{F} = 6.23 \times 10^{-3} \, \text{N}$ (in the positive x-direction)

Step 1: Break Down the Forces

Since the net force is in the positive x-direction, the magnetic force must cancel out the negative y-component created by the magnetic field. Therefore, we can express this relationship as:

$F_x = qE + q(v_y B)$

Where $v_y$ is the y-component of the velocity.

Step 2: Set Up the Equations

1. The x-component of the force: $6.23 \times 10^{-3} = q(1450) + q(v_y \cdot 1.22)$

2. The y-component of the force (since the net force in y-direction must be zero): $0 = q(v_x \cdot 1.22) - q(v_y \cdot 1450)$

Step 3: Solve for Velocity Components

1. From the second equation: $v_x \cdot 1.22 = v_y \cdot 1450 \quad \Rightarrow \quad v_y = \frac{1.22}{1450} v_x$

2. Substitute $v_y$ into the first equation: $6.23 \times 10^{-3} = q(1450) + q\left(\frac{1.22}{1450} v_x \cdot 1.22\right)$

   Simplifying: $6.23 \times 10^{-3} = q\left(1450 + \frac{1.4884}{1450} v_x\right)$

Step 4: Isolate $v_x$

For simplicity, let's assume $q$ is a constant that can be factored out (this will cancel later). We can express the net force equation in terms of $v_x$.

Assuming $q$ does not change the relative terms: $6.23 \times 10^{-3} = q \cdot 1450 + \frac{1.4884}{1450} q v_x$

Step 5: Rearrange and Solve

We can solve for $v_x$ as follows:

1. Rearranging gives: $6.23 \times 10^{-3} - 1450q = \frac{1.4884}{1450} q v_x$

2. Factor out $q$: $v_x = \frac{(6.23 \times 10^{-3} - 1450q) \cdot 1450}{1.4884}$

Final Step: Magnitude and Direction

At this point, we need the value of $q$ to compute $v_x$ and subsequently $v_y$. However, you can find the relationship of the magnitude:

$v = \sqrt{v_x^2 + v_y^2}$ $\tan(\theta) = \frac{v_y}{v_x}$

Summary

Without a specific value for $q$, we can express the relationship for the components and solve for $v_x$ and $v_y$ once $q$ is known.

Would you like to proceed with specific numerical calculations or further explore the theoretical aspects?

Follow-Up Questions:

1. What is the charge of the particle?
2. Do you need clarification on how to derive the components?
3. Would you like to see an example with specific values?
4. How does the direction of velocity relate to the forces involved?
5. What is the significance of the angle of the velocity vector in this context?

Tip: Understanding the direction of the magnetic force (which is perpendicular to both the velocity and magnetic field) is key to solving problems involving charged particles in electromagnetic fields.