Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m`
A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces.
The speed with which the particle will collide the screen is

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem step by step, we will analyze the forces acting on the charged particle and calculate its speed when it collides with the screen.

### Step 1: Understand the Forces Acting on the Particle
The particle has a charge \( q = 1 \, \text{C} \) and mass \( m = 1 \, \text{kg} \). It is released from rest at the origin, and it will experience an electric force due to the electric field \( \vec{E} \) and a magnetic force due to the magnetic field \( \vec{B} \).

### Step 2: Calculate the Electric Force
The electric field is given by:
\[ \vec{E} = 10 \hat{i} \, \text{N/C} \]
The electric force \( \vec{F}_e \) acting on the particle is given by:
\[ \vec{F}_e = q \vec{E} = 1 \times (10 \hat{i}) = 10 \hat{i} \, \text{N} \]

### Step 3: Determine the Acceleration
Using Newton's second law, the acceleration \( \vec{a} \) of the particle can be calculated as:
\[ \vec{a} = \frac{\vec{F}_e}{m} = \frac{10 \hat{i}}{1} = 10 \hat{i} \, \text{m/s}^2 \]

### Step 4: Calculate the Velocity at \( x = 1.8 \, \text{m} \)
The particle moves under constant acceleration from rest. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity),
- \( a = 10 \, \text{m/s}^2 \) (acceleration),
- \( s = 1.8 \, \text{m} \) (distance traveled).

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 1.8 \]
\[ v^2 = 36 \implies v = 6 \, \text{m/s} \]

### Step 5: Analyze Motion in the Magnetic Field
At \( x = 1.8 \, \text{m} \), the particle enters a region with a magnetic field given by:
\[ \vec{B} = -5 \hat{k} \, \text{T} \]
Since the velocity \( \vec{v} = 6 \hat{i} \, \text{m/s} \) is perpendicular to the magnetic field, the particle will undergo circular motion due to the magnetic force.

### Step 6: Determine the Speed During Circular Motion
In uniform circular motion, the speed of the particle remains constant. Since the particle enters the magnetic field with a speed of \( 6 \, \text{m/s} \), it will continue to move at this speed while in the magnetic field.

### Step 7: Conclusion
When the particle collides with the screen at \( x = 3 \, \text{m} \), it will have the same speed it had when it entered the magnetic field, which is:
\[ \text{Speed at collision} = 6 \, \text{m/s} \]

Thus, the speed with which the particle will collide with the screen is **6 m/s**.

---