A proton charge (+ e coulomb) enters in a magnetic field of strength B (Tesla) perpendicular to the magnetic lines of force, with speed v. The force on the proton is -

Options

A. evB
B. 0
C. `infty`
D. `evB//2`

Answer

To find the force acting on a proton when it enters a magnetic field perpendicular to its velocity, we can use the formula for the magnetic force on a charged particle. Here’s the step-by-step solution:

### Step 1: Identify the formula for magnetic force
The magnetic force \( F \) acting on a charged particle moving in a magnetic field is given by the equation:
\[ F = Q \cdot v \cdot B \cdot \sin(\theta) \]
where:
- \( F \) is the magnetic force,
- \( Q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step 2: Substitute known values
In this case, the proton has a charge \( Q = +e \) (where \( e \approx 1.6 \times 10^{-19} \, \text{C} \)), the velocity is \( v \), and the magnetic field strength is \( B \). The angle \( \theta \) is \( 90^\circ \) because the velocity is perpendicular to the magnetic field lines.

### Step 3: Calculate the sine of the angle
Since \( \theta = 90^\circ \):
\[ \sin(90^\circ) = 1 \]

### Step 4: Simplify the equation
Substituting \( \sin(90^\circ) \) into the formula gives:
\[ F = Q \cdot v \cdot B \cdot 1 \]
Thus, we can simplify it to:
\[ F = Q \cdot v \cdot B \]

### Step 5: Substitute the charge of the proton
Now, substituting \( Q = e \):
\[ F = e \cdot v \cdot B \]

### Final Answer
The force acting on the proton is:
\[ F = e \cdot v \cdot B \]

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