A uniform magnetic field `B = 1.2` mT is directed vertically upward throughout the volume of a laboratory chamber . A proton (`m_(p) = 1.67 xx 10^(-27)` kg) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is `3 xx 10^(7)` m/s

Options

A. `3.45 xx 10^(12) m//s^(2)`
B. `1.67 xx 10^(12) m//s^(2)`
C. `5.25 xx 10^(12) m//s^(2)`
D. `2.75 xx 10^(12) m//s^(2)`

Explanation

To solve the problem of calculating the magnitude of centripetal acceleration of a proton moving in a magnetic field, we will follow these steps:

### Step 1: Understand the Given Information
We have the following information:
- Magnetic field strength, \( B = 1.2 \, \text{mT} = 1.2 \times 10^{-3} \, \text{T} \)
- Mass of the proton, \( m_p = 1.67 \times 10^{-27} \, \text{kg} \)
- Speed of the proton, \( v = 3 \times 10^7 \, \text{m/s} \)

### Step 2: Calculate the Charge of the Proton
The charge of a proton is a known constant:
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 3: Determine the Centripetal Force
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F = qvB \]
This force acts as the centripetal force required to keep the proton moving in a circular path. Therefore, we can equate the magnetic force to the centripetal force:
\[ F = \frac{mv^2}{r} \]
Where \( r \) is the radius of the circular path.

### Step 4: Relate the Forces
Setting the magnetic force equal to the centripetal force, we have:
\[ qvB = \frac{mv^2}{r} \]

### Step 5: Solve for Centripetal Acceleration
Centripetal acceleration \( a_c \) can be expressed as:
\[ a_c = \frac{v^2}{r} \]
From the previous equation, we can express \( r \) in terms of \( q, v, B, \) and \( m \):
\[ r = \frac{mv}{qB} \]
Substituting \( r \) into the centripetal acceleration formula gives:
\[ a_c = \frac{v^2}{\frac{mv}{qB}} = \frac{qvB}{m} \]

### Step 6: Substitute the Values
Now we can substitute the known values into the equation:
\[ a_c = \frac{(1.6 \times 10^{-19} \, \text{C})(3 \times 10^7 \, \text{m/s})(1.2 \times 10^{-3} \, \text{T})}{1.67 \times 10^{-27} \, \text{kg}} \]

### Step 7: Calculate the Result
Calculating the numerator:
\[ 1.6 \times 10^{-19} \times 3 \times 10^7 \times 1.2 \times 10^{-3} = 5.76 \times 10^{-15} \]
Now, dividing by the mass of the proton:
\[ a_c = \frac{5.76 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 3.44 \times 10^{12} \, \text{m/s}^2 \]

### Final Answer
The magnitude of the centripetal acceleration of the proton is:
\[ \boxed{3.44 \times 10^{12} \, \text{m/s}^2} \]