The force on a charged particle due to electric and magnetic fields is given by `vecF=qvecE+qvecvXvecB`. Suppose `vecE` is along the X-axis and `vecB` along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?

Explanation

To solve the problem, we need to find the direction and minimum speed \( v \) of a positively charged particle such that the net force on it due to electric and magnetic fields is zero. The force on the charged particle is given by:

\[ \vec{F} = q\vec{E} + q\vec{v} \times \vec{B} \]

### Step 1: Understand the Direction of Fields
Given:
- The electric field \( \vec{E} \) is along the X-axis.
- The magnetic field \( \vec{B} \) is along the Y-axis.

We can represent these fields as:
\[ \vec{E} = E \hat{i} \]
\[ \vec{B} = B \hat{j} \]

### Step 2: Set Up the Force Equation
For the net force to be zero, we have:
\[ \vec{F} = 0 \implies q\vec{E} + q\vec{v} \times \vec{B} = 0 \]
This simplifies to:
\[ \vec{E} = -\vec{v} \times \vec{B} \]

### Step 3: Express the Velocity Vector
Let the velocity vector \( \vec{v} \) be expressed in terms of its components:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \]

### Step 4: Calculate the Cross Product
Now we compute \( \vec{v} \times \vec{B} \):
\[ \vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times (B \hat{j}) \]
Using the right-hand rule and properties of cross products, we find:
\[ \vec{v} \times \vec{B} = v_x B \hat{k} - v_y B \hat{i} \]

### Step 5: Set Up the Equation
Substituting this into our earlier equation gives:
\[ \vec{E} = - (v_x B \hat{k} - v_y B \hat{i}) \]
This leads to:
\[ \vec{E} = v_y B \hat{i} - v_x B \hat{k} \]

### Step 6: Compare Components
Now we can compare components from both sides:
1. The \( \hat{i} \) component gives:
\[ E = v_y B \]
2. The \( \hat{k} \) component gives:
\[ 0 = -v_x B \implies v_x = 0 \]

### Step 7: Solve for \( v_y \) and \( v_z \)
Since \( v_x = 0 \), we can substitute this back into the equation for \( E \):
\[ E = v_y B \]
From this, we can express \( v_y \):
\[ v_y = \frac{E}{B} \]

### Step 8: Determine the Direction and Minimum Speed
Since \( v_x = 0 \) and \( v_y = \frac{E}{B} \), the velocity vector can be expressed as:
\[ \vec{v} = 0 \hat{i} + \frac{E}{B} \hat{j} + v_z \hat{k} \]
To ensure the net force is zero, we need to have a component along the \( z \)-axis:
\[ v_z = \frac{E}{B} \]

### Final Result
The minimum speed \( v \) of the charged particle should be:
\[ v = \sqrt{v_y^2 + v_z^2} = \sqrt{\left(\frac{E}{B}\right)^2 + \left(\frac{E}{B}\right)^2} = \frac{E}{B} \sqrt{2} \]
The direction of the velocity vector is in the \( z \)-direction.