A proton of mass m and charge q is moving in a plane with kinetic energy E. if there exists a uniform magnetic field B, perpendicular to the plane motion. The proton will move in a circular path of radius

Options

A. `(3Em)/(qB)`
B. `(sqrt(2Em))/(qB)`
C. `(sqrt(Em))/(2qB)`
D. `(sqrt(2Eq))/(qB)`

Answer

To find the radius of the circular path of a proton moving in a magnetic field, we can follow these steps:

### Step 1: Understand the Forces Acting on the Proton
When a charged particle like a proton moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force required for circular motion. The magnetic force \( F_B \) is given by:
\[ F_B = qvB \]
where:
- \( q \) is the charge of the proton,
- \( v \) is the velocity of the proton,
- \( B \) is the magnetic field strength.

### Step 2: Set Up the Equation for Circular Motion
For an object moving in a circular path, the centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{R} \]
where:
- \( m \) is the mass of the proton,
- \( R \) is the radius of the circular path.

### Step 3: Equate the Forces
Since the magnetic force provides the necessary centripetal force, we can set the two forces equal to each other:
\[ qvB = \frac{mv^2}{R} \]

### Step 4: Solve for the Radius \( R \)
Rearranging the equation to solve for \( R \):
\[ R = \frac{mv}{qB} \]

### Step 5: Relate Velocity to Kinetic Energy
The kinetic energy \( E \) of the proton is related to its mass and velocity by the equation:
\[ E = \frac{1}{2}mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2E}{m}} \]

### Step 6: Substitute Velocity into the Radius Equation
Now, substituting \( v \) back into the equation for \( R \):
\[ R = \frac{m \sqrt{\frac{2E}{m}}}{qB} \]
This simplifies to:
\[ R = \frac{\sqrt{2mE}}{qB} \]

### Final Answer
Thus, the radius of the circular path of the proton is given by:
\[ R = \frac{\sqrt{2mE}}{qB} \]