A charged particle of charge q and mass m is shot into a uniform magnetic field of induction B at an angle `theta` with the field. The frequency of revolution of the particle

Options

A. Depends on the angle `theta`
B. Is independent of the angle `theta`
C. Is proportional to the specific charge q/m of the particle
D. Is inversely proportional to the value of B

Answer

To find the frequency of revolution of a charged particle moving in a magnetic field at an angle \(\theta\), we can follow these steps:

### Step 1: Understand the motion of the charged particle
When a charged particle with charge \(q\) and mass \(m\) enters a magnetic field of induction \(B\) at an angle \(\theta\), it experiences a magnetic force that acts perpendicular to its velocity. This force causes the particle to move in a circular path.

### Step 2: Determine the effective velocity component
The velocity \(v\) of the charged particle can be decomposed into two components:
- The component parallel to the magnetic field: \(v_{\parallel} = v \cos(\theta)\)
- The component perpendicular to the magnetic field: \(v_{\perpendicular} = v \sin(\theta)\)

Only the perpendicular component contributes to the circular motion in the magnetic field.

### Step 3: Apply the formula for the frequency of revolution
The frequency \(f\) of the circular motion of a charged particle in a magnetic field is given by the formula:
\[ f = \frac{qB}{2\pi m} \]
However, since only the perpendicular component of the velocity contributes to the circular motion, we need to consider the effective velocity \(v_{\perpendicular}\).

### Step 4: Relate the frequency to the perpendicular velocity
The centripetal force required for circular motion is provided by the magnetic force:
\[ F = qv_{\perpendicular}B \]
The centripetal force can also be expressed as:
\[ F = \frac{mv_{\perpendicular}^2}{r} \]
Setting these two expressions for force equal gives:
\[ qv_{\perpendicular}B = \frac{mv_{\perpendicular}^2}{r} \]
From this, we can derive the radius \(r\) of the circular path:
\[ r = \frac{mv_{\perpendicular}}{qB} \]

### Step 5: Calculate the frequency using the effective velocity
Now, substituting \(v_{\perpendicular} = v \sin(\theta)\) into the frequency formula:
\[ f = \frac{qB}{2\pi m} \cdot \frac{v \sin(\theta)}{r} \]
Since the radius \(r\) is already expressed in terms of \(v_{\perpendicular}\), we can simplify the expression for frequency:
\[ f = \frac{qB \sin(\theta)}{2\pi m} \]

### Final Expression
Thus, the frequency of revolution of the charged particle in the magnetic field at an angle \(\theta\) is given by:
\[ f = \frac{qB \sin(\theta)}{2\pi m} \]

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