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. `(5hati+5sqrt3hatj)m//s`
B. `10hatJm//s`
C. `(5sqrt3hati-5hatj)m//s`
D. `-10hatjm//s`

Answer

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

### Step 1: Understand the motion of the particle
The particle is projected with an initial velocity \( \vec{V}_0 = 10 \, \text{m/s} \) along the positive x-axis. The magnetic field is given as \( \vec{B} = -2 \hat{k} \) Tesla, which means it is directed along the negative z-axis.

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

### Step 3: Calculate the magnetic force
The magnetic force acting on the particle can be calculated using the formula:
\[ \vec{F} = q (\vec{V} \times \vec{B}) \]
Since the particle is initially moving in the x-direction and the magnetic field is in the z-direction, the force will act perpendicular to both the velocity and the magnetic field.

### Step 4: Determine the angular velocity
The angular velocity \( \omega \) of the particle in circular motion due to the magnetic field is given by:
\[ \omega = \frac{qB}{m} \]
Substituting \( \frac{q}{m} = \pi \) and \( B = 2 \):
\[ \omega = \pi \times 2 = 2\pi \, \text{rad/s} \]

### Step 5: Calculate the time period
The time period \( T \) of the circular motion can be calculated as:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2\pi} = 1 \, \text{s} \]

### Step 6: Calculate the angle rotated in \( t = \frac{1}{6} \) s
The angle \( \theta \) rotated in time \( t \) is given by:
\[ \theta = \omega t = 2\pi \times \frac{1}{6} = \frac{\pi}{3} \, \text{radians} \]

### Step 7: Determine the new velocity components
The initial velocity vector is:
\[ \vec{V}_0 = 10 \hat{i} \, \text{m/s} \]
After rotating through an angle of \( \frac{\pi}{3} \) (or 60 degrees), the new velocity components can be calculated using trigonometric functions:
\[ \vec{V} = |\vec{V}_0| \left( \cos\left(\frac{\pi}{3}\right) \hat{i} + \sin\left(\frac{\pi}{3}\right) \hat{j} \right) \]
Calculating the components:
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)

Thus,
\[ \vec{V} = 10 \left( \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) = 5 \hat{i} + 5\sqrt{3} \hat{j} \, \text{m/s} \]

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