CUET · PHYSICS · PYQ PAPER 2025
The frequency of revolution of an electron revolving in a circular path in a magnetic field of \( 7.8 \times 10^{-4} \, \text{T} \) will be (Take \( e = -1.6 \times 10^{-19} \, \text{C} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \), \( \pi = 22/7 \))
- A 42 MHz
- B 21.8 MHz
- C 220 MHz
- D 46 MHz
Answer & Solution
Correct Answer
(B) 21.8 MHz
Step-by-step Solution
Detailed explanation
\( f = \frac{|e|B}{2\pi m_e} \) \( f = \frac{(1.6 \times 10^{-19} \, \text{C}) \times (7.8 \times 10^{-4} \, \text{T})}{2 \times (22/7) \times (9.1 \times 10^{-31} \, \text{kg})} \) \( f = 2.18 \times 10^7 \, \text{Hz} = 21.8 \, \text{MHz} \)
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