A particle of specific charge `q/m = pi C kg^-1` is projected from the origin toward positive x-axis with a velocity of `10 ms^-1` in a uniform magnetic field `vec B = -2 hat k T.` The velocity `vec v` of particle after time `t=1/12 s` will be (in `ms^-1`)

Options

A. `5[hati +sqrt(3)hatj]`
B. `5[sqrt(3)hati +hatj]`
C. `5[sqrt(3)hati -sqrt(3)hatj]`
D. `5[hati +hatj]`

Answer

To solve the problem step by step, we will analyze the motion of the particle in the magnetic field and calculate its velocity after a given time.

### Step 1: Understand the Forces Acting on the Particle
The particle has a specific charge \( \frac{q}{m} = \pi \, \text{C/kg} \) and is projected with an initial velocity \( \vec{v} = 10 \, \text{m/s} \) along the positive x-axis. The magnetic field is given by \( \vec{B} = -2 \hat{k} \, \text{T} \). The magnetic force acting on the particle is given by the Lorentz force equation:

\[ \vec{F} = q \vec{v} \times \vec{B} \]

### Step 2: Calculate the Magnetic Force
Since \( \vec{v} \) is along the x-axis and \( \vec{B} \) is along the negative z-axis, we can calculate the cross product:

\[ \vec{F} = q \vec{v} \times \vec{B} = q (10 \hat{i}) \times (-2 \hat{k}) \]

Using the right-hand rule, we find:

\[ \vec{F} = q \cdot 10 \cdot (-2) \hat{i} \times \hat{k} = -20q \hat{j} \]

### Step 3: Determine the Angular Speed
The magnetic force acts as a centripetal force, causing the particle to move in a circular path. The angular speed \( \omega \) can be calculated using the relationship:

\[ \omega = \frac{qB}{m} \]

Substituting \( \frac{q}{m} = \pi \) and \( B = 2 \):

\[ \omega = \pi \cdot 2 = 2\pi \, \text{rad/s} \]

### Step 4: Calculate the Angle Rotated in Time \( t = \frac{1}{12} \, \text{s} \)
The angle \( \theta \) rotated in time \( t \) can be calculated using:

\[ \theta = \omega t = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} \, \text{rad} \quad \text{(or 30 degrees)} \]

### Step 5: Determine the New Velocity Vector
Initially, the velocity vector is:

\[ \vec{v}_0 = 10 \hat{i} \]

After rotating by \( \frac{\pi}{6} \) radians, we can find the new velocity components using trigonometric functions:

\[ v_x = v_0 \cos(\theta) = 10 \cdot \cos\left(\frac{\pi}{6}\right) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

\[ v_y = v_0 \sin(\theta) = 10 \cdot \sin\left(\frac{\pi}{6}\right) = 10 \cdot \frac{1}{2} = 5 \]

Thus, the new velocity vector after time \( t \) is:

\[ \vec{v} = 5\sqrt{3} \hat{i} + 5 \hat{j} \]

### Step 6: Final Answer
The velocity of the particle after \( t = \frac{1}{12} \, \text{s} \) is:

\[ \vec{v} = 5\sqrt{3} \hat{i} + 5 \hat{j} \, \text{m/s} \]