A charged particle with velocity `hatv=xhati+yhatj` moves in a magnetic field `vecB=yhati+xhatj`. The force acting on the particle has magnitude F. Which one of the following statements is are correct?

Options

A. No force will act on particle, if x=y
B. `Fprop(x^(2)-y^(2))`if x is greater than y
C. The force will act along z-axis, if x greater than y
D. The force will act along y-axis, if y greater than x

Answer

To solve the problem step-by-step, we need to analyze the given information about the charged particle's velocity and the magnetic field, and then apply the formula for the magnetic force.

### Step 1: Identify the velocity and magnetic field vectors
The velocity of the charged particle is given as:
\[ \vec{v} = x \hat{i} + y \hat{j} \]
The magnetic field is given as:
\[ \vec{B} = y \hat{i} + x \hat{j} \]

### Step 2: Use the formula for magnetic force
The magnetic force \(\vec{F}\) acting on a charged particle moving in a magnetic field is given by the equation:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
where \(q\) is the charge of the particle.

### Step 3: Calculate the cross product \(\vec{v} \times \vec{B}\)
To find \(\vec{F}\), we need to calculate the cross product \(\vec{v} \times \vec{B}\):
\[ \vec{v} \times \vec{B} = (x \hat{i} + y \hat{j}) \times (y \hat{i} + x \hat{j}) \]

Using the properties of the cross product:
- \(\hat{i} \times \hat{i} = 0\)
- \(\hat{j} \times \hat{j} = 0\)
- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{i} = -\hat{k}\)

We can expand the cross product:
\[ \vec{v} \times \vec{B} = x \hat{i} \times y \hat{i} + x \hat{i} \times x \hat{j} + y \hat{j} \times y \hat{i} + y \hat{j} \times x \hat{j} \]
This simplifies to:
\[ = 0 + x y \hat{k} + 0 - y x \hat{k} \]
Thus:
\[ \vec{v} \times \vec{B} = (x^2 - y^2) \hat{k} \]

### Step 4: Write the expression for the magnetic force
Now substituting back into the force equation:
\[ \vec{F} = q (x^2 - y^2) \hat{k} \]

### Step 5: Analyze the magnitude and direction of the force
The magnitude of the magnetic force is:
\[ F = |q| |x^2 - y^2| \]
- If \(x = y\), then \(F = 0\) (no force acts on the particle).
- If \(x > y\), then \(F\) is positive, and the force acts in the positive \(z\)-direction.
- If \(y > x\), then \(F\) is negative, and the force acts in the negative \(z\)-direction.

### Conclusion
Based on this analysis, we can evaluate the provided options:
- **Option A**: Correct, no force acts if \(x = y\).
- **Option B**: Correct, force is proportional to \(x^2 - y^2\) when \(x > y\).
- **Option C**: Correct, the force acts along the \(z\)-axis if \(x > y\).
- **Option D**: Incorrect, the force does not act along the \(y\)-axis if \(y > x\).

### Final Answer
The correct statements are A, B, and C.

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