A proton of energy `2 MeV` is moving perpendicular to uniform magnetic field of `2.5T`. The form on the proton is `(Mp=1.6xx10^(-27)Kg` and `q=e=1.6xx10^(-19)C)`

Options

A. `2.5xx10^(-10)N`
B. `7.6xx10^(-11)N`
C. `2.5xx10^(-11)N`
D. `8xx10^(-12)N`

Answer

To solve the problem of finding the force on a proton moving perpendicular to a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Energy of the proton, \( E = 2 \, \text{MeV} = 2 \times 10^6 \, \text{eV} \)
- Magnetic field strength, \( B = 2.5 \, \text{T} \)
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton, \( M_p = 1.6 \times 10^{-27} \, \text{kg} \)

### Step 2: Convert energy from electron volts to joules
To convert the energy from MeV to joules, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ E = 2 \times 10^6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-13} \, \text{J} \]

### Step 3: Relate kinetic energy to velocity
The kinetic energy (KE) of the proton can be expressed as:
\[ KE = \frac{1}{2} M_p v^2 \]
Setting this equal to the energy we calculated:
\[ 3.2 \times 10^{-13} = \frac{1}{2} (1.6 \times 10^{-27}) v^2 \]

### Step 4: Solve for velocity \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2 \times 3.2 \times 10^{-13}}{1.6 \times 10^{-27}} = \frac{6.4 \times 10^{-13}}{1.6 \times 10^{-27}} = 4 \times 10^{14} \]
Taking the square root to find \( v \):
\[ v = \sqrt{4 \times 10^{14}} = 2 \times 10^7 \, \text{m/s} \]

### Step 5: Calculate the magnetic force
The magnetic force \( F \) on a charged particle moving in a magnetic field is given by:
\[ F = q v B \]
Substituting the known values:
\[ F = (1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^7 \, \text{m/s}) \times (2.5 \, \text{T}) \]
Calculating this:
\[ F = 1.6 \times 2 \times 2.5 \times 10^{-19} \times 10^7 = 8 \times 10^{-12} \, \text{N} \]

### Final Answer
The force on the proton is:
\[ F = 8 \times 10^{-12} \, \text{N} \]

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