CUET · PHYSICS · PYQ PAPER 2025
An electron placed in a magnetic field of \( 3.14 \times 10^{-4} \, \text{T} \) moves in a circle. The frequency of the revolution of the electron is:
- A \( 8.79 \times 10^6 \, \text{Hz} \)
- B \( 8.79 \times 10^5 \, \text{Hz} \)
- C \( 6.08 \times 10^6 \, \text{Hz} \)
- D \( 2.8 \times 10^6 \, \text{Hz} \)
Answer & Solution
Correct Answer
(A) \( 8.79 \times 10^6 \, \text{Hz} \)
Step-by-step Solution
Detailed explanation
\( f = \frac{eB}{2\pi m_e} \) \( f = \frac{(1.602 \times 10^{-19} \, \text{C}) \times (3.14 \times 10^{-4} \, \text{T})}{2\pi \times (9.109 \times 10^{-31} \, \text{kg})} \) \( f = 8.79 \times 10^6 \, \text{Hz} \)
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