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. 0.71mT
B. 71mT
C. 7.1mT
D. 0.071mT

Answer

To solve the problem, we will follow these steps:

### Step 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.

\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \]

### Step 2: Calculate the Velocity of the Proton
The kinetic energy of the proton can be expressed using the formula:

\[ KE = \frac{1}{2} m v^2 \]

Where:
- \( KE = 1.6 \times 10^{-13} \text{ J} \)
- \( m \) (mass of the proton) \( = 1.67 \times 10^{-27} \text{ kg} \)

Rearranging the formula to find \( v \):

\[ v^2 = \frac{2 \times KE}{m} \]

Substituting the values:

\[ v^2 = \frac{2 \times 1.6 \times 10^{-13}}{1.67 \times 10^{-27}} \]

Calculating \( v^2 \):

\[ v^2 = \frac{3.2 \times 10^{-13}}{1.67 \times 10^{-27}} \approx 1.916 \times 10^{14} \]

Taking the square root to find \( v \):

\[ v \approx \sqrt{1.916 \times 10^{14}} \approx 1.38 \times 10^7 \text{ m/s} \]

### Step 3: Use the Formula for Magnetic Force
The magnetic force acting on the proton can be expressed as:

\[ F = q v B \]

Where:
- \( q \) (charge of the proton) \( = 1.6 \times 10^{-19} \text{ C} \)
- \( a \) (acceleration) \( = 10^{12} \text{ m/s}^2 \)

From Newton's second law, we know:

\[ F = m a \]

Setting the two expressions for force equal to each other:

\[ m a = q v B \]

### Step 4: Solve for the Magnetic Field \( B \)
Rearranging the equation to solve for \( B \):

\[ B = \frac{m a}{q v} \]

Substituting the known values:

\[ B = \frac{(1.67 \times 10^{-27} \text{ kg}) (10^{12} \text{ m/s}^2)}{(1.6 \times 10^{-19} \text{ C}) (1.38 \times 10^7 \text{ m/s})} \]

Calculating \( B \):

\[ B \approx \frac{1.67 \times 10^{-15}}{2.208 \times 10^{-12}} \approx 0.000757 \text{ T} = 0.757 \text{ mT} \]

### Step 5: Convert to milliTesla
Converting to milliTesla:

\[ B \approx 0.757 \text{ mT} \approx 0.71 \text{ mT} \]

### Final Answer
The magnitude of the magnetic field is approximately:

\[ B \approx 0.71 \text{ mT} \]

---