A particle of specific charge `q//m = (pi) C//kg` is projected from the origin towards positive x-axis with a velocity of `10 m//s` in a uniform magnetic field `vec(B) = -2 hat K` Tesla. The velocity `vec V` of the particle after time `t =1//6` s will be

Options

A. `(5 hat(i)+5sqrt(3) hat(j))m//s`
B. `10 hat(j)m//s`
C. `(5 sqrt(3)hat(i)+5hat(j))m//s`
D. `-10 hat(j) m//s`

Explanation

To solve the problem, we need to determine the velocity of a particle after a specific time when it is subjected to a magnetic field. Here is the step-by-step solution:

### Step 1: Understand the initial conditions
The particle is projected from the origin towards the positive x-axis with an initial velocity \( \vec{V_0} = 10 \, \text{m/s} \) and the magnetic field is given by \( \vec{B} = -2 \hat{k} \, \text{T} \).

### Step 2: Determine the specific charge
The specific charge of the particle is given as \( \frac{q}{m} = \pi \, \text{C/kg} \).

### Step 3: Analyze the motion in the magnetic field
Since the velocity of the particle is perpendicular to the magnetic field, the particle will undergo circular motion due to the Lorentz force. The radius of the circular motion and the time period can be calculated using the specific charge and the magnetic field.

### Step 4: Calculate the time period
The time period \( T \) of the circular motion is given by the formula:
\[ T = \frac{2\pi m}{qB} \]
Substituting \( \frac{q}{m} = \pi \) and \( B = 2 \):
\[ T = \frac{2\pi}{\pi \cdot 2} = 1 \, \text{s} \]

### Step 5: Determine the angular displacement after \( t = \frac{1}{6} \, \text{s} \)
In 1 second, the particle completes one full rotation (360 degrees). Therefore, in \( \frac{1}{6} \, \text{s} \), the angular displacement \( \theta \) is:
\[ \theta = \frac{1}{6} \times 360^\circ = 60^\circ \]

### Step 6: Calculate the new velocity components
The new velocity components can be determined using the angle of rotation. The components of the velocity after \( t = \frac{1}{6} \, \text{s} \) can be expressed as:
\[ \vec{V} = V \cos(60^\circ) \hat{i} + V \sin(60^\circ) \hat{j} \]
Where \( V = 10 \, \text{m/s} \).

### Step 7: Substitute the values of cosine and sine
Using \( \cos(60^\circ) = \frac{1}{2} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ \vec{V} = 10 \left( \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) \]
\[ \vec{V} = 5 \hat{i} + 5\sqrt{3} \hat{j} \, \text{m/s} \]

### Final Answer
The velocity of the particle after \( t = \frac{1}{6} \, \text{s} \) is:
\[ \vec{V} = 5 \hat{i} + 5\sqrt{3} \hat{j} \, \text{m/s} \]

---