An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 imes 10

Question

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

Accepted Answer

1. **Identify the Physical Relationship:** For a charged particle moving in a circular path perpendicular to a uniform magnetic field $B$, the magnetic force provides the centripetal force. The angular frequency is $\omega = \frac{qB}{m}$ .
2. **Frequency Formula:** The frequency of revolution (cyclotron frequency) is given by $
u = \frac{\omega}{2\pi} = \frac{1}{2\pi} \left( \frac{e}{m} ight) B$ .
3. **Substitute Given Values:** 
   - $B = 3.57 	imes 10^{-2}$ T
   - $e/m = 1.76 	imes 10^{11}$ C/kg
   - $
u = \frac{1}{2 	imes 3.14159} 	imes (1.76 	imes 10^{11}) 	imes (3.57 	imes 10^{-2})$
4. **Calculation:**
   - $
u = \frac{6.2832 	imes 10^{9}}{6.28318} \approx 1 	imes 10^{9}$ Hz
5. **Unit Conversion:** $10^9$ Hz is equal to 1 GHz. Therefore, the frequency of revolution is 1 GHz.

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