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.
y-coordinate of particle where it collides with screen (in meters) is

Options

A. `(0.6(sqrt3-1))/(sqrt3)`
B. `(0.6(sqrt3+1))/(sqrt3)`
C. `1.2(sqrt3+1)`
D. `(1.2(sqrt3-1))/(sqrt3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the acceleration of the charged particle due to the electric field.

The electrostatic force on the charged particle is given by:

\[ \vec{F}_e = q \vec{E} \]

Given:
- Charge \( q = 1 \, \text{C} \)
- Electric field \( \vec{E} = 10 \hat{i} \, \text{N/C} \)

The force becomes:

\[ \vec{F}_e = 1 \cdot (10 \hat{i}) = 10 \hat{i} \, \text{N} \]

Now, using Newton's second law, we can find the acceleration \( \vec{a} \):

\[ \vec{a} = \frac{\vec{F}_e}{m} = \frac{10 \hat{i}}{1} = 10 \hat{i} \, \text{m/s}^2 \]

### Step 2: Calculate the velocity of the particle when it reaches \( x = 1.8 \, \text{m} \).

Using the kinematic equation:

\[ v^2 = u^2 + 2as \]

Where:
- Initial velocity \( u = 0 \)
- Acceleration \( a = 10 \, \text{m/s}^2 \)
- Distance \( s = 1.8 \, \text{m} \)

Substituting the values:

\[ v^2 = 0 + 2 \cdot 10 \cdot 1.8 = 36 \]

Thus, the velocity \( v \) is:

\[ v = \sqrt{36} = 6 \, \text{m/s} \]

### Step 3: Analyze the motion of the particle in the magnetic field.

At \( x = 1.8 \, \text{m} \), the particle enters a magnetic field \( \vec{B} = -5 \hat{k} \, \text{T} \). The magnetic force \( \vec{F}_m \) is given by:

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

Substituting the values:

\[ \vec{v} = 6 \hat{i} \, \text{m/s}, \quad \vec{B} = -5 \hat{k} \, \text{T} \]

Calculating the cross product:

\[ \vec{v} \times \vec{B} = (6 \hat{i}) \times (-5 \hat{k}) = -30 \hat{j} \, \text{N} \]

Thus, the magnetic force is:

\[ \vec{F}_m = 1 \cdot (-30 \hat{j}) = -30 \hat{j} \, \text{N} \]

### Step 4: Calculate the radius of the circular motion.

The magnetic force provides the centripetal force:

\[ F_m = \frac{mv^2}{r} \]

Setting the forces equal:

\[ 30 = \frac{1 \cdot (6)^2}{r} \]

Solving for \( r \):

\[ 30 = \frac{36}{r} \implies r = \frac{36}{30} = 1.2 \, \text{m} \]

### Step 5: Determine the distance the particle travels in the magnetic field.

The particle travels from \( x = 1.8 \, \text{m} \) to \( x = 2.4 \, \text{m} \):

\[ \Delta x = 2.4 - 1.8 = 0.6 \, \text{m} \]

### Step 6: Find the angle of deflection.

Using the geometry of the circular motion, we can find the angle \( \theta \):

\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{0.6}{1.2} = 0.5 \]

Thus,

\[ \theta = 30^\circ \]

### Step 7: Determine the y-coordinate when the particle strikes the screen at \( x = 3 \, \text{m} \).

The additional distance traveled in the x-direction is:

\[ \Delta x' = 3 - 2.4 = 0.6 \, \text{m} \]

Using the tangent function:

\[ \tan(30^\circ) = \frac{y}{0.6} \]

Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):

\[ \frac{1}{\sqrt{3}} = \frac{y}{0.6} \implies y = 0.6 \cdot \frac{1}{\sqrt{3}} = \frac{0.6}{\sqrt{3}} \approx 0.346 \, \text{m} \]

### Final Result

The y-coordinate of the particle when it collides with the screen is approximately:

\[ y \approx 0.346 \, \text{m} \]