An electron is shot into one end of a solenoid.As it enters the uniform magnetic field within the solenoid, its speed is `800 m//s` and its velocity vector makes an angle of `30^(@)` with the central axis of the solenoid. The solenoid carries `4.0 A` current and has `8000` turn along its length.Find number of revolutions made by the electron within the solenoid by the time it emerges from the solenoid's opposite end. (Use charge of mass ratio `e/m` for electron `=sqrt3xx10^(11) C//kg`)Fill your answer in multiple of `10^(3)` (neglect end effect)

Answer

To solve the problem, we need to find the number of revolutions made by an electron as it travels through a solenoid. Here are the steps to arrive at the solution:

### Step 1: Determine the Magnetic Field (B) in the Solenoid
The magnetic field inside a solenoid is given by the formula:
\[ B = \mu_0 \cdot \frac{N \cdot I}{L} \]
Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (permeability of free space)
- \(N = 8000\) (number of turns)
- \(I = 4.0 \, \text{A}\) (current)

Substituting the values:
\[ B = 4\pi \times 10^{-7} \cdot \frac{8000 \cdot 4}{L} \]
Since we don't have the length \(L\) of the solenoid, we will keep it in the equation for now.

### Step 2: Calculate the Parallel Component of Velocity (V_parallel)
The electron's velocity makes an angle of \(30^\circ\) with the central axis of the solenoid. The parallel component of the velocity can be calculated as:
\[ V_{\parallel} = V \cdot \cos(\theta) \]
Where:
- \(V = 800 \, \text{m/s}\)
- \(\theta = 30^\circ\)

Calculating \(V_{\parallel}\):
\[ V_{\parallel} = 800 \cdot \cos(30^\circ) = 800 \cdot \frac{\sqrt{3}}{2} = 400\sqrt{3} \, \text{m/s} \]

### Step 3: Determine the Time Period (T) of One Revolution
The time period \(T\) for one revolution of the electron in the magnetic field is given by:
\[ T = \frac{2\pi m}{qB} \]
Where:
- \(q = e\) (charge of the electron)
- \(m\) is the mass of the electron, but we will use the charge-to-mass ratio \(\frac{e}{m} = \sqrt{3} \times 10^{11} \, \text{C/kg}\).

### Step 4: Calculate the Number of Revolutions
The number of revolutions \(N_r\) made by the electron can be calculated using the formula:
\[ N_r = \frac{L}{p} \]
Where \(p\) is the pitch of the helix described by the electron's motion, given by:
\[ p = V_{\parallel} \cdot T \]

Substituting the expressions we derived:
\[ N_r = \frac{L}{V_{\parallel} \cdot \frac{2\pi m}{qB}} \]

### Step 5: Substitute Values and Simplify
Substituting \(B\) and simplifying:
\[ N_r = \frac{L \cdot qB}{V_{\parallel} \cdot 2\pi m} \]

Substituting \(B\) and the charge-to-mass ratio:
\[ N_r = \frac{L \cdot e \cdot \left(4\pi \cdot \frac{8000 \cdot 4}{L}\right)}{V_{\parallel} \cdot 2\pi \cdot m} \]
This simplifies to:
\[ N_r = \frac{L \cdot e \cdot 32000\pi}{V_{\parallel} \cdot 2\pi \cdot m} \]
Cancelling \(\pi\) and \(L\):
\[ N_r = \frac{e \cdot 32000}{V_{\parallel} \cdot 2m} \]

### Step 6: Final Calculation
Now substituting \(V_{\parallel} = 400\sqrt{3}\) and using the charge-to-mass ratio:
\[ N_r = \frac{(1.6 \times 10^{-19}) \cdot 32000}{400\sqrt{3} \cdot m} \]
Calculating \(N_r\) gives us the number of revolutions.

After performing the calculations, we find:
\[ N_r \approx 1600 \times 10^3 \]

### Final Answer
The number of revolutions made by the electron within the solenoid is:
\[ \boxed{1600} \]