A proton is projected with a speed of `3X10^6 ms ^(-1)` horizontally from east to west. A uniform magnetic field `vec B` of strength `2.0X10^(-3)` T exists in the vertically upward direction(a) find the force on the proton just after it is projected. (b) what is the acceleration produced?

Options

A. `11.6 xx 10^(11) m//s^(2)`
B. `17.4xx10^(11)m//s`
C. `5.8xx10^(11)m//s^(2)`
D. `2.9xx10^(11)m//s^(2)`

Answer

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Finding the Force on the Proton

1. **Identify the Given Values**:
- Speed of the proton, \( v = 3 \times 10^6 \, \text{m/s} \)
- Magnetic field strength, \( B = 2.0 \times 10^{-3} \, \text{T} \)
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- The direction of the velocity is from east to west (horizontally).
- The direction of the magnetic field is vertically upward.

2. **Use the Formula for Magnetic Force**:
The force \( F \) on a charged particle moving in a magnetic field is given by the equation:
\[ F = q \cdot (v \times B) \]
Since the velocity \( v \) and the magnetic field \( B \) are perpendicular to each other, we can simplify this to:
\[ F = q \cdot v \cdot B \]

3. **Substitute the Values**:
\[ F = (1.6 \times 10^{-19} \, \text{C}) \cdot (3 \times 10^6 \, \text{m/s}) \cdot (2.0 \times 10^{-3} \, \text{T}) \]

4. **Calculate the Force**:
\[ F = 1.6 \times 3 \times 2.0 \times 10^{-19} \times 10^6 \times 10^{-3} \]
\[ F = 9.6 \times 10^{-16} \, \text{N} \]

### Part (b): Finding the Acceleration Produced

1. **Use Newton's Second Law**:
The acceleration \( a \) produced on the proton can be calculated using Newton's second law:
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the proton.

2. **Identify the Mass of the Proton**:
The mass of the proton is given as:
\[ m = 1.67 \times 10^{-27} \, \text{kg} \]

3. **Substitute the Values**:
\[ a = \frac{9.6 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} \]

4. **Calculate the Acceleration**:
\[ a = 5.74 \times 10^{11} \, \text{m/s}^2 \]

### Final Answers:
- (a) The force on the proton is \( 9.6 \times 10^{-16} \, \text{N} \).
- (b) The acceleration produced is approximately \( 5.74 \times 10^{11} \, \text{m/s}^2 \).