A charge particle moves with velocity `vecV=Ahati+Dhatj` magnetic field . `vecB=Ahati+Dhatj` The force acting on the particle has magnitude .F Then,

Options

A. F = 0 , if aD = dA
B. F = 0 , if aD = -dA
C. F = 0 , if aA = -dD
D. `F prop (a^2+b^2)^(1//2)xx(A^2+D^(2))^(1//2)`

Explanation

To solve the problem, we need to find the force acting on a charged particle moving in a magnetic field. The force can be calculated using the formula:

\[ \vec{F} = Q (\vec{v} \times \vec{B}) \]

Where:
- \( Q \) is the charge of the particle,
- \( \vec{v} \) is the velocity vector of the particle,
- \( \vec{B} \) is the magnetic field vector.

Given:
- \( \vec{v} = A \hat{i} + D \hat{j} \)
- \( \vec{B} = A \hat{i} + D \hat{j} \)

### Step 1: Write down the vectors
We have:
- Velocity vector: \( \vec{v} = A \hat{i} + D \hat{j} \)
- Magnetic field vector: \( \vec{B} = A \hat{i} + D \hat{j} \)

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)
To find the cross product, we can use the determinant method:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A & D & 0 \\ A & D & 0 \end{vmatrix} \]

### Step 3: Expand the determinant
Using the determinant formula:

\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} D & 0 \\ D & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} A & 0 \\ A & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} A & D \\ A & D \end{vmatrix} \]

Calculating each of the determinants:
- The first determinant (for \( \hat{i} \)): \( D \cdot 0 - D \cdot 0 = 0 \)
- The second determinant (for \( \hat{j} \)): \( A \cdot 0 - A \cdot 0 = 0 \)
- The third determinant (for \( \hat{k} \)): \( A \cdot D - A \cdot D = 0 \)

So, we have:

\[ \vec{v} \times \vec{B} = 0 \hat{i} - 0 \hat{j} + 0 \hat{k} = \vec{0} \]

### Step 4: Calculate the force \( \vec{F} \)
Now substituting back into the force equation:

\[ \vec{F} = Q (\vec{v} \times \vec{B}) = Q \cdot \vec{0} = \vec{0} \]

### Conclusion
The force acting on the charged particle is zero.