A magnetic field `vec(B)=-B_(0)hat(i)` exists within a sphere of radius `R=v_(0)Tsqrt(3)` where T is the time period of one revolution of a charged particle starting its motion form origin and moving with a velocity `vce(v)_(0) = (v_0)/(2) sqrt(3) hat(i) - v_(0)/(2) hat(j)`. Find the number of turns that the particle will take to come out of the magnetic field.

Explanation

To solve the problem, we need to analyze the motion of a charged particle moving in a magnetic field. The magnetic field is given as \(\vec{B} = -B_0 \hat{i}\), and the particle's velocity is given as:

\[ \vec{v} = \frac{v_0}{2} \sqrt{3} \hat{i} - \frac{v_0}{2} \hat{j} \]

### Step 1: Identify the components of velocity
The velocity can be broken down into two components:
- The component in the \(\hat{i}\) direction: \(v_{x} = \frac{v_0}{2} \sqrt{3}\)
- The component in the \(\hat{j}\) direction: \(v_{y} = -\frac{v_0}{2}\)

### Step 2: Calculate the magnitude of the velocity
The magnitude of the velocity \(v\) can be calculated using the Pythagorean theorem:

\[ v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{\left(\frac{v_0}{2} \sqrt{3}\right)^2 + \left(-\frac{v_0}{2}\right)^2} \]

Calculating this gives:

\[ v = \sqrt{\frac{3v_0^2}{4} + \frac{v_0^2}{4}} = \sqrt{\frac{4v_0^2}{4}} = v_0 \]

### Step 3: Determine the radius of the circular motion
The radius \(R\) of the circular motion in the magnetic field can be expressed as:

\[ R = \frac{mv}{qB} \]

However, we are given that \(R = v_0 T \sqrt{3}\).

### Step 4: Calculate the time period of one revolution
The time period \(T\) is the time taken for one complete revolution. The particle moves in a helical path due to the velocity components. The component of velocity perpendicular to the magnetic field contributes to circular motion, while the component parallel to the magnetic field contributes to linear motion.

The perpendicular component of the velocity \(v_{\perp}\) is:

\[ v_{\perp} = v_y = -\frac{v_0}{2} \]

### Step 5: Calculate the pitch of the helical path
The pitch \(P\) of the helical path is given by the distance traveled in the direction of the magnetic field during one complete revolution. This is calculated as:

\[ P = v_{\parallel} \cdot T \]

Where \(v_{\parallel}\) is the component of the velocity parallel to the magnetic field. Here, \(v_{\parallel} = v_x = \frac{v_0}{2} \sqrt{3}\).

### Step 6: Calculate the number of turns
The number of turns \(n\) the particle makes while it is within the magnetic field can be calculated using the formula:

\[ n = \frac{R}{P} \]

Substituting the values we have:

\[ n = \frac{v_0 T \sqrt{3}}{v_{\parallel} \cdot T} = \frac{v_0 T \sqrt{3}}{\frac{v_0}{2} \sqrt{3} \cdot T} \]

The \(T\) and \(v_0\) cancel out:

\[ n = \frac{\sqrt{3}}{\frac{1}{2} \sqrt{3}} = 2 \]

### Final Answer
The number of turns that the particle will take to come out of the magnetic field is \(n = 2\).

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