A proton moves with a speed of `2.0times10^(7)ms^(-1)` along the x -axis.It enters a region where there is a magnetic field of magnitude `1.5 T` directed at an angle of `30^(@)` to the x -axis and lying in the xy plane.The magnitude of the magnetic force on the proton is

Options

A. `24xx10^(-12)N`
B. `2.4xx10^(-12)N`
C. `8.0xx10^(-12)N`
D. `0.8xx10^(-12)N`

Answer

To find the magnitude of the magnetic force acting on a proton moving in a magnetic field, we can use the formula for the magnetic force:

\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the proton,
- \( v \) is the speed of the proton,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step 1: Identify the given values
- Speed of the proton, \( v = 2.0 \times 10^7 \, \text{m/s} \)
- Magnetic field strength, \( B = 1.5 \, \text{T} \)
- Angle, \( \theta = 30^\circ \)
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate \( \sin(\theta) \)
Since \( \theta = 30^\circ \):
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula:
\[ F = (1.6 \times 10^{-19} \, \text{C}) \cdot (2.0 \times 10^7 \, \text{m/s}) \cdot (1.5 \, \text{T}) \cdot \sin(30^\circ) \]

### Step 4: Perform the calculation
Substituting \( \sin(30^\circ) = \frac{1}{2} \):
\[ F = (1.6 \times 10^{-19}) \cdot (2.0 \times 10^7) \cdot (1.5) \cdot \left(\frac{1}{2}\right) \]

Calculating step-by-step:
1. Calculate \( 1.6 \times 10^{-19} \cdot 2.0 \times 10^7 \):
\[ = 3.2 \times 10^{-12} \]
2. Multiply by \( 1.5 \):
\[ = 3.2 \times 10^{-12} \cdot 1.5 = 4.8 \times 10^{-12} \]
3. Finally, multiply by \( \frac{1}{2} \):
\[ F = \frac{4.8 \times 10^{-12}}{2} = 2.4 \times 10^{-12} \, \text{N} \]

### Final Answer
The magnitude of the magnetic force on the proton is:
\[ F = 2.4 \times 10^{-12} \, \text{N} \]