An electron is moving in a circular path under the influence fo a transerve magnetic field of `3.57xx10^(-2)T`. If the value of `e//m` is `1.76xx10^(141) C//kg`. The frequency of revolution of the electron is

Options

A. `1GH z`
B. `100 MHz`
C. `62.8 MHz`
D. `6.28MHz`

Answer

To find the frequency of revolution of the electron moving in a circular path under the influence of a transverse magnetic field, we can follow these steps:

### Step 1: Understand the relationship between magnetic force and centripetal force
When a charged particle, such as an electron, 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 electron,
- \( V \) is the velocity of the electron,
- \( B \) is the magnetic field strength.

The centripetal force \( F_C \) required to keep the electron in circular motion is given by:
\[ F_C = \frac{MV^2}{R} \]
where:
- \( M \) is the mass of the electron,
- \( R \) is the radius of the circular path.

### Step 2: Set the magnetic force equal to the centripetal force
Since the magnetic force provides the centripetal force, we can set them equal to each other:
\[ QVB = \frac{MV^2}{R} \]

### Step 3: Solve for the radius \( R \)
Rearranging the equation gives:
\[ R = \frac{MV}{QB} \]

### Step 4: Determine the time period \( T \)
The time period \( T \) of one complete revolution is given by:
\[ T = \frac{2\pi R}{V} \]
Substituting the expression for \( R \):
\[ T = \frac{2\pi \left(\frac{MV}{QB}\right)}{V} = \frac{2\pi M}{QB} \]

### Step 5: Calculate the frequency \( f \)
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{QB}{2\pi M} \]

### Step 6: Substitute known values
We know:
- The charge of the electron \( Q = 1.6 \times 10^{-19} \, C \),
- The magnetic field \( B = 3.57 \times 10^{-2} \, T \),
- The value of \( \frac{e}{m} = 1.76 \times 10^{11} \, \frac{C}{kg} \).

Using \( f = \frac{e}{m} \cdot \frac{B}{2\pi} \):
\[ f = \left(1.76 \times 10^{11} \, \frac{C}{kg}\right) \cdot \left(\frac{3.57 \times 10^{-2} \, T}{2\pi}\right) \]

### Step 7: Calculate the frequency
Calculating the numerical values:
\[ f = 1.76 \times 10^{11} \cdot \frac{3.57 \times 10^{-2}}{2 \cdot 3.14} \]
\[ f \approx 1.76 \times 10^{11} \cdot \frac{3.57 \times 10^{-2}}{6.28} \approx 1.76 \times 10^{11} \cdot 5.68 \times 10^{-3} \approx 10^{9} \, Hz \]

### Final Answer
The frequency of revolution of the electron is approximately \( 1 \, GHz \).

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