A proton enter in a uniform magnetic field of 2.0 mT at an angle of 60degrees with the magnetic field with speed 10m/s. Find the picth of path

Options

A. 30pi microm
B. 50pi microm
C. 80pi microm
D. 10pi microm

Answer

To solve the problem of finding the pitch of the path of a proton entering a uniform magnetic field, we can follow these steps:

### Step 1: Understand the Given Data
- Magnetic field \( B = 2.0 \, \text{mT} = 2.0 \times 10^{-3} \, \text{T} \)
- Speed of the proton \( v = 10 \, \text{m/s} \)
- Angle with the magnetic field \( \theta = 60^\circ \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \)

### Step 2: Calculate the Radius of the Circular Path
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force. The formula for the radius \( r \) of the circular path is given by:

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

Substituting the known values:

\[ r = \frac{(1.67 \times 10^{-27} \, \text{kg})(10 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C})(2.0 \times 10^{-3} \, \text{T})} \]

Calculating the denominator:

\[ qB = (1.6 \times 10^{-19})(2.0 \times 10^{-3}) = 3.2 \times 10^{-22} \]

Now substituting back to find \( r \):

\[ r = \frac{(1.67 \times 10^{-27})(10)}{3.2 \times 10^{-22}} = \frac{1.67 \times 10^{-26}}{3.2 \times 10^{-22}} \approx 5.22 \times 10^{-5} \, \text{m} \]

### Step 3: Calculate the Time Period of Circular Motion
The time period \( T \) for one complete revolution in a circular path is given by:

\[ T = \frac{2\pi r}{v} \]

Substituting the values:

\[ T = \frac{2\pi (5.22 \times 10^{-5})}{10} = \frac{2\pi \times 5.22 \times 10^{-5}}{10} \approx 3.28 \times 10^{-5} \, \text{s} \]

### Step 4: Calculate the Pitch of the Path
The pitch \( P \) of the helical path is given by the formula:

\[ P = v \cos(\theta) \cdot T \]

Calculating \( v \cos(60^\circ) \):

\[ \cos(60^\circ) = \frac{1}{2} \Rightarrow v \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

Now substituting into the pitch formula:

\[ P = (5 \, \text{m/s}) \cdot (3.28 \times 10^{-5} \, \text{s}) = 1.64 \times 10^{-4} \, \text{m} \]

### Step 5: Convert to Micrometers
To convert meters to micrometers:

\[ P = 1.64 \times 10^{-4} \, \text{m} = 164 \, \mu m \]

### Final Answer
The pitch of the path of the proton is approximately **164 micrometers**.

---