A proton is projected with a speed of `2 xx 10^(5)` m/s at an angle `60^(circ)` to `x` -axis. If a uniform magnetic field of 0.1 T is 'applied along `y` -axis, then the path of proton is helical with time period `a times 10^(-5) s . (Take `pi=3.14)`

Answer

To solve the problem, we need to find the time period of the helical motion of a proton projected in a magnetic field. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the given values
- Speed of the proton, \( v = 2 \times 10^5 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Magnetic field strength, \( B = 0.1 \, \text{T} \)
- Charge of the proton, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton, \( m = 1.67 \times 10^{-27} \, \text{kg} \)

### Step 2: Calculate the component of velocity perpendicular to the magnetic field
The velocity component perpendicular to the magnetic field is given by:
\[ v_{\perp} = v \sin \theta \]
Substituting the values:
\[ v_{\perp} = 2 \times 10^5 \sin(60^\circ) = 2 \times 10^5 \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^5 \, \text{m/s} \]

### Step 3: Calculate the radius of the helical path
The radius \( r \) of the helical path is given by:
\[ r = \frac{mv_{\perp}}{eB} \]
Substituting the values:
\[ r = \frac{(1.67 \times 10^{-27} \, \text{kg}) \cdot (\sqrt{3} \times 2 \times 10^5 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \cdot (0.1 \, \text{T})} \]

### Step 4: Calculate the time period of the helical motion
The time period \( T \) of the helical motion is given by:
\[ T = \frac{2\pi r}{v_{\perp}} \]
Substituting the expression for \( r \):
\[ T = \frac{2\pi}{eB} \cdot \frac{m}{v_{\perp}} \]
Substituting the known values:
\[ T = \frac{2 \cdot 3.14 \cdot (1.67 \times 10^{-27})}{(1.6 \times 10^{-19}) \cdot (0.1)} \]

### Step 5: Simplify the expression
Calculating the numerical values:
\[ T = \frac{2 \cdot 3.14 \cdot 1.67 \times 10^{-27}}{1.6 \times 10^{-20}} = \frac{10.56 \times 10^{-27}}{1.6 \times 10^{-20}} = 6.6 \times 10^{-7} \, \text{s} \]

### Step 6: Convert to the desired form
To express \( T \) in the form \( a \times 10^{-5} \):
\[ T = 0.066 \times 10^{-5} \, \text{s} = 6.6 \times 10^{-6} \, \text{s} \]

### Final Answer
The time period \( T \) is approximately \( 6.6 \times 10^{-6} \, \text{s} \).