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

### Step 1: Identify the velocity and magnetic field vectors
Given:
\[ \vec{v} = a \hat{i} + d \hat{j} \]
\[ \vec{B} = A \hat{i} + D \hat{j} \]

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)
To find the force, we need to calculate the cross product \( \vec{v} \times \vec{B} \).

Using the determinant form for the cross product:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & d & 0 \\ A & D & 0 \end{vmatrix} \]

### Step 3: Expand the determinant
Calculating the determinant, we have:
\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} d & 0 \\ D & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} a & 0 \\ A & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} a & d \\ A & D \end{vmatrix} \]

Calculating the minors:
- The first minor gives \( d \cdot 0 - 0 \cdot D = 0 \)
- The second minor gives \( a \cdot 0 - 0 \cdot A = 0 \)
- The third minor gives \( aD - dA \)

Thus:
\[ \vec{v} \times \vec{B} = 0 \hat{i} - 0 \hat{j} + (aD - dA) \hat{k} = (aD - dA) \hat{k} \]

### Step 4: Calculate the magnitude of the force
The magnitude of the force \( F \) is given by:
\[ F = q |\vec{v} \times \vec{B}| = q |(aD - dA) \hat{k}| \]
\[ F = q |aD - dA| \]

### Step 5: Determine when the force is zero
The force \( F \) will be zero when:
\[ aD - dA = 0 \]
This implies:
\[ aD = dA \]
or rearranging,
\[ \frac{a}{A} = \frac{d}{D} \]

### Conclusion
Therefore, the force acting on the particle is zero if the above condition holds true.