A proton is moving perpendicular to a uniform magnetic field of 2.5 tesla with 2 MeV kinetic energy. The force on proton is _______N. (Mass of proton= `1.6 xx 10^(-27)` kg, charge of proton `=1.6 xx 10^(-19)` C)

Options

A. `8 xx 10^(-12)`
B. `8 xx 10^(-11)`
C. `3 xx 10^(-11)`
D. `3 xx 10^(-10)`

Answer

To find the force on a proton moving perpendicular to a magnetic field, we can use the formula for magnetic force:

\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the proton,
- \( v \) is the velocity of the proton,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity and the magnetic field.

### Step 1: Convert the kinetic energy of the proton to velocity
The kinetic energy (KE) of the proton is given as 2 MeV. We need to convert this to joules:

1 MeV = \( 1.6 \times 10^{-13} \) Joules

Thus,

\[ KE = 2 \, \text{MeV} = 2 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-13} \, \text{J} \]

Using the kinetic energy formula:

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

We can rearrange this to solve for \( v \):

\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

Substituting the values:

- Mass of the proton, \( m = 1.6 \times 10^{-27} \, \text{kg} \)
- Kinetic energy, \( KE = 3.2 \times 10^{-13} \, \text{J} \)

\[ v = \sqrt{\frac{2 \cdot 3.2 \times 10^{-13}}{1.6 \times 10^{-27}}} \]

Calculating this:

\[ v = \sqrt{\frac{6.4 \times 10^{-13}}{1.6 \times 10^{-27}}} = \sqrt{4 \times 10^{14}} = 2 \times 10^{7} \, \text{m/s} \]

### Step 2: Calculate the force using the magnetic force formula
Now that we have the velocity, we can calculate the magnetic force.

Given:
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field strength, \( B = 2.5 \, \text{T} \)
- Velocity, \( v = 2 \times 10^{7} \, \text{m/s} \)
- Since the proton is moving perpendicular to the magnetic field, \( \sin(90^\circ) = 1 \).

Now substituting these values into the force formula:

\[ F = q \cdot v \cdot B \]

\[ F = (1.6 \times 10^{-19}) \cdot (2 \times 10^{7}) \cdot (2.5) \]

Calculating this:

\[ F = 1.6 \times 2 \times 2.5 \times 10^{-19} \times 10^{7} \]

\[ F = 8 \times 10^{-12} \, \text{N} \]

### Final Answer:
The force on the proton is \( 8 \times 10^{-12} \, \text{N} \).

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