A charged particle moves with velocity `vec v = a hat i + d hat j` in a magnetic field `vec B = A hat i + D hat j.` 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. `fprop (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 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: Set up the cross product
To find the force, we need to compute the cross product \( \vec{v} \times \vec{B} \). The cross product can be calculated using the determinant of a matrix:

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

### Step 3: Calculate the determinant
Calculating the determinant, we find:

\[ \vec{v} \times \vec{B} = \hat{i}(d \cdot 0 - 0 \cdot D) - \hat{j}(a \cdot 0 - 0 \cdot A) + \hat{k}(aD - dA) \]

This simplifies to:

\[ \vec{v} \times \vec{B} = (aD - dA) \hat{k} \]

### Step 4: Write the force vector
Now substituting back into the force equation:

\[ \vec{F} = Q \left( (aD - dA) \hat{k} \right) \]

### Step 5: Find the magnitude of the force
The magnitude of the force \( F \) is given by:

\[ F = |Q| |aD - dA| \]

### Step 6: Analyze the condition for zero force
For the force to be zero, we need:

\[ aD - dA = 0 \]

This implies:

\[ aD = dA \]

### Conclusion
Thus, the condition for the force to be zero is:

\[ aD = dA \quad \text{or} \quad \frac{a}{d} = \frac{A}{D} \quad \text{(if } d \neq 0 \text{ and } D \neq 0\text{)} \]