When proton of KE = 1.0 MeV moving in South to North direction enters the magnetic field (from West to East direction), it accelerates with `a = 10^(12) m/s^2` . The magnitude of magnetic field is-

Options

A. `0.71 mT`
B. `7.1 mT`
C. `71 mT`
D. `710 mT`

Explanation

To find the magnitude of the magnetic field when a proton with a kinetic energy of 1.0 MeV accelerates with an acceleration of \( a = 10^{12} \, \text{m/s}^2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Kinetic Energy to Joules**:
The kinetic energy (KE) of the proton is given as 1.0 MeV. We need to convert this to joules:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Therefore,
\[ KE = 1.0 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]

2. **Calculate the Velocity of the Proton**:
The kinetic energy is related to the velocity \( v \) by the equation:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
The mass \( m \) of a proton is approximately \( 1.67 \times 10^{-27} \, \text{kg} \). Plugging in the values:
\[ v = \sqrt{\frac{2 \cdot 1.6 \times 10^{-13} \, \text{J}}{1.67 \times 10^{-27} \, \text{kg}}} \]

3. **Calculate the Velocity**:
Performing the calculation:
\[ v = \sqrt{\frac{3.2 \times 10^{-13}}{1.67 \times 10^{-27}}} = \sqrt{1.916 \times 10^{14}} \approx 1.38 \times 10^7 \, \text{m/s} \]

4. **Use the Formula for Magnetic Force**:
The magnetic force \( F \) acting on a charged particle moving in a magnetic field is given by:
\[ F = qvB \]
where \( q \) is the charge of the proton (approximately \( 1.6 \times 10^{-19} \, \text{C} \)), \( v \) is the velocity, and \( B \) is the magnetic field strength.

5. **Set the Magnetic Force Equal to the Centripetal Force**:
The magnetic force also acts as the centripetal force, which can be expressed as:
\[ F = ma \]
Therefore, we can set:
\[ qvB = ma \]
Rearranging for \( B \):
\[ B = \frac{ma}{qv} \]

6. **Substitute the Values**:
Now substituting the known values:
\[ B = \frac{(1.67 \times 10^{-27} \, \text{kg})(10^{12} \, \text{m/s}^2)}{(1.6 \times 10^{-19} \, \text{C})(1.38 \times 10^7 \, \text{m/s})} \]

7. **Calculate the Magnetic Field**:
Performing the calculation:
\[ B = \frac{1.67 \times 10^{-15}}{2.208 \times 10^{-12}} \approx 0.756 \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field \( B \) is approximately \( 0.756 \, \text{T} \).