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 magnetic field, we can follow these steps:

### Step 1: Understand the Given Data
- Magnetic field strength, \( B = 2.0 \, \text{mT} = 2 \times 10^{-3} \, \text{T} \)
- Speed of the proton, \( v = 10 \, \text{m/s} \)
- Angle with the magnetic field, \( \theta = 60^\circ \)

### Step 2: Calculate the Radius of the Circular Path
The force acting on the proton due to the magnetic field (Lorentz force) provides the centripetal force required for circular motion. The Lorentz force is given by:
\[ F = qvB \sin(\theta) \]
And the centripetal force is given by:
\[ F = \frac{mv^2}{r} \]
Setting these two forces equal gives us:
\[ qvB \sin(\theta) = \frac{mv^2}{r} \]
Rearranging for the radius \( r \):
\[ r = \frac{mv}{qB \sin(\theta)} \]

### Step 3: Calculate \( \frac{m}{q} \)
For a proton:
- Mass \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)

Calculating \( \frac{m}{q} \):
\[ \frac{m}{q} = \frac{1.67 \times 10^{-27}}{1.6 \times 10^{-19}} \approx 1.04 \times 10^{-8} \, \text{kg/C} \]

### Step 4: Substitute Values into the Radius Formula
Substituting \( \frac{m}{q} \) into the radius formula:
\[ r = \frac{(1.04 \times 10^{-8}) \cdot 10}{(2 \times 10^{-3}) \cdot \sin(60^\circ)} \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ r = \frac{(1.04 \times 10^{-8}) \cdot 10}{(2 \times 10^{-3}) \cdot \frac{\sqrt{3}}{2}} = \frac{(1.04 \times 10^{-7})}{(2 \times 10^{-3}) \cdot \frac{\sqrt{3}}{2}} = \frac{(1.04 \times 10^{-7})}{(10^{-3} \sqrt{3})} \]
Calculating this gives:
\[ r \approx \frac{1.04 \times 10^{-4}}{\sqrt{3}} \approx 6.0 \times 10^{-5} \, \text{m} \]

### Step 5: Calculate the Pitch of the Path
The pitch \( P \) of the helical path is given by:
\[ P = v \cos(\theta) \cdot T \]
Where \( T \) is the time period of one complete revolution:
\[ T = \frac{2\pi r}{v} \]
Substituting \( T \):
\[ P = v \cos(\theta) \cdot \frac{2\pi r}{v} \]
This simplifies to:
\[ P = 2\pi r \cos(\theta) \]
Substituting \( r \) and \( \cos(60^\circ) = \frac{1}{2} \):
\[ P = 2\pi (6.0 \times 10^{-5}) \cdot \frac{1}{2} = \pi \times 6.0 \times 10^{-5} \approx 1.88 \times 10^{-4} \, \text{m} = 188 \, \mu\text{m} \]

### Final Answer
The pitch of the path is approximately \( 188 \, \mu\text{m} \).

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