Question

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2} T$. If the value of $e / m$ is $1.76 \times 10^{11} C / kg$, the frequency of revolution of the electron is

• A. 1 GHz
• B. 100 MHz
• C. 62.8 MHz
• D. 6.28 MHz

Answer 1

The Correct Option is A

Frequency Calculation 

The formula for frequency \( f \) is given by:

\(f = \frac{qB}{2 \pi m}\)

Where: - \( q \) is the charge, - \( B \) is the magnetic field strength, - \( m \) is the mass of the particle.

Rearranging the formula, we get:

\(f = \left( \frac{q}{m} \right) \times \frac{B}{2 \pi}\)

Substituting the given values: - \( q = 1.76 \times 10^{11} \, \text{C/kg} \), - \( B = 3.57 \times 10^{-2} \, \text{T} \), - \( m = 1 \, \text{kg} \),

\(f = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{-2}}{2 \times 3.14}\)

Simplifying the expression:

\(f = 1 \times 10^9 \, \text{Hz}\)

Finally, converting to GHz:

\(f = 1 \, \text{GHz}\)

Conclusion:

The frequency \( f \) is \( 1 \, \text{GHz} \).