CUET · PHYSICS · PYQ PAPER 2023
What is the radius of the path of an electron moving with velocity of \(3 \times 10^7\) m/s in a magnetic field of \(6 \times 10^{-4}\) T perpendicular to it?
- A 14 cm
- B 28 cm
- C 56 cm
- D 112 cm
Answer & Solution
Correct Answer
(B) 28 cm
Step-by-step Solution
Detailed explanation
\(r = \frac{mv}{qB}\) \(r = \frac{(9.11 \times 10^{-31} \text{ kg})(3 \times 10^7 \text{ m/s})}{(1.602 \times 10^{-19} \text{ C})(6 \times 10^{-4} \text{ T})}\) \(r \approx 0.284 \text{ m} \approx 28 \text{ cm}\)
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