A charged particle enters a uniform magnetic field with velocity `v_(0) = 4 m//s` perpendicular to it, the length of magnetic field is `x = ((sqrt(3))/(2)) R`, where R is the radius of the circular path of the particle in the field. Find the magnitude of charge in velocity (in m/s) of the particle when it comes out of the field.

Explanation

To solve the problem step by step, we will analyze the motion of a charged particle in a magnetic field and derive the final velocity when it exits the field.

### Step 1: Understanding the Motion
When a charged particle enters a uniform magnetic field perpendicularly, it moves in a circular path due to the magnetic force acting on it. The radius \( R \) of the circular path is determined by the particle's velocity \( v_0 \) and the magnetic field \( B \).

### Step 2: Given Parameters
- Initial velocity \( v_0 = 4 \, \text{m/s} \)
- Length of the magnetic field \( x = \frac{\sqrt{3}}{2} R \)

### Step 3: Finding the Radius of the Circular Path
The radius \( R \) of the circular path can be expressed in terms of the magnetic field \( B \) and the charge \( q \) of the particle using the formula:
\[ R = \frac{mv_0}{qB} \]
where \( m \) is the mass of the particle.

### Step 4: Finding the Angle
From the given length of the magnetic field \( x \), we can relate it to the angle \( \theta \) subtended by the circular path:
\[ x = R \theta \]
Given \( x = \frac{\sqrt{3}}{2} R \), we can write:
\[ R \theta = \frac{\sqrt{3}}{2} R \]
This implies:
\[ \theta = \frac{\sqrt{3}}{2} \]
To find \( \theta \) in degrees, we can convert:
\[ \theta = 60^\circ \]

### Step 5: Velocity Components
When the particle exits the magnetic field, we can resolve the velocity into components. The new velocity \( v_1 \) can be expressed as:
\[ v_1 = v_0 \cos(60^\circ) \hat{i} + v_0 \sin(60^\circ) \hat{j} \]
Substituting \( v_0 = 4 \, \text{m/s} \):
\[ v_1 = 4 \cdot \frac{1}{2} \hat{i} + 4 \cdot \frac{\sqrt{3}}{2} \hat{j} \]
This simplifies to:
\[ v_1 = 2 \hat{i} + 2\sqrt{3} \hat{j} \]

### Step 6: Magnitude of the Final Velocity
To find the magnitude of the final velocity \( v_1 \):
\[ |v_1| = \sqrt{(2)^2 + (2\sqrt{3})^2} \]
Calculating this gives:
\[ |v_1| = \sqrt{4 + 12} = \sqrt{16} = 4 \, \text{m/s} \]

### Final Answer
The magnitude of the velocity of the particle when it comes out of the field is \( 4 \, \text{m/s} \).

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