A charged particle of charge q and mass m enters perpendiculalry in a magnetic field B. Kinetic energy of particle E, then frequency of rotation is

Options

A. `(qB)/(mpi)`
B. `qB/(2pim)`
C. `(qBE)/(2pim)`
D. `(qB)/(2piE)`

Answer

To find the frequency of rotation of a charged particle with charge \( q \) and mass \( m \) entering perpendicularly into a magnetic field \( B \), we can follow these steps:

### Step 1: Determine the Magnetic Force
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F = q(v \times B) \]
Since the particle enters the magnetic field perpendicularly, the magnitude of this force simplifies to:
\[ F = qvB \]

### Step 2: Relate Magnetic Force to Centripetal Force
As the charged particle moves in a circular path due to the magnetic force, this force acts as the centripetal force required to keep the particle in circular motion. The centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( r \) is the radius of the circular path.

### Step 3: Set the Forces Equal
Setting the magnetic force equal to the centripetal force, we have:
\[ qvB = \frac{mv^2}{r} \]

### Step 4: Solve for Velocity
Rearranging the equation to solve for \( v \):
\[ qvB = \frac{mv^2}{r} \implies qB = \frac{mv}{r} \implies v = \frac{qBr}{m} \]

### Step 5: Determine Angular Velocity
The angular velocity \( \omega \) is related to the linear velocity \( v \) and the radius \( r \) by the equation:
\[ v = r\omega \]
Substituting the expression for \( v \):
\[ \frac{qBr}{m} = r\omega \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ \omega = \frac{qB}{m} \]

### Step 6: Calculate Frequency
The frequency \( f \) of rotation is related to the angular velocity \( \omega \) by:
\[ \omega = 2\pi f \]
Substituting for \( \omega \):
\[ 2\pi f = \frac{qB}{m} \]
Now, solving for \( f \):
\[ f = \frac{qB}{2\pi m} \]

### Final Answer
Thus, the frequency of rotation of the charged particle is:
\[ f = \frac{qB}{2\pi m} \]

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