An electron is projected at an angle `theta` with a uniform magnetic field. If the pitch of the helical path is equal to its radius, then the angle of projection is

Options

A. (a)`tan^(-1)pi`
B. (b)`tan^(-1)2pi`
C. (c)`cot^(-1)pi`
D. (d)`cot^(-1)2pi`

Explanation

To solve the problem step by step, we need to analyze the conditions given and derive the necessary relationships.

### Step 1: Understand the problem
We have an electron projected at an angle θ with respect to a uniform magnetic field. The problem states that the pitch of the helical path is equal to its radius.

### Step 2: Define the radius of the helical path
The radius \( r \) of the helical path of a charged particle moving in a magnetic field is given by the formula:
\[ r = \frac{mv \sin \theta}{qB} \]
where:
- \( m \) = mass of the electron
- \( v \) = velocity of the electron
- \( \theta \) = angle of projection
- \( q \) = charge of the electron
- \( B \) = magnetic field strength

### Step 3: Define the pitch of the helical path
The pitch \( p \) of the helical path is given by:
\[ p = v \cos \theta \cdot T \]
where \( T \) is the time period of the motion. The time period \( T \) for circular motion in a magnetic field is:
\[ T = \frac{2\pi m}{qB} \]
Thus, substituting for \( T \):
\[ p = v \cos \theta \cdot \frac{2\pi m}{qB} \]

### Step 4: Set the radius equal to the pitch
According to the problem, we have:
\[ r = p \]
Substituting the expressions for \( r \) and \( p \):
\[ \frac{mv \sin \theta}{qB} = v \cos \theta \cdot \frac{2\pi m}{qB} \]

### Step 5: Simplify the equation
We can cancel \( v \) (assuming \( v \neq 0 \)), \( m \), and \( qB \) from both sides:
\[ \sin \theta = 2\pi \cos \theta \]

### Step 6: Rearrange the equation
Dividing both sides by \( \cos \theta \) gives:
\[ \tan \theta = 2\pi \]

### Step 7: Find the angle of projection
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}(2\pi) \]

### Conclusion
Thus, the angle of projection \( \theta \) is:
\[ \theta = \tan^{-1}(2\pi) \]