CUET · PHYSICS · PYQ PAPER 2025
A proton is projected with a speed of \(4 \times 10^6 \text{ m s}^{-1}\) horizontally from east to west. A uniform magnetic field of strength \(3 \times 10^{-3} \text{ T}\) exists in the vertically upward direction. The acceleration of the proton is:
(Given: mass of proton \(1.67 \times 10^{-27} \text{ kg}\); charge on proton \(1.6 \times 10^{-19} \text{ C}\))
- A zero
- B \(11.5 \times 10^{11} \text{ m s}^{-2}\)
- C \(18.3 \times 10^{10} \text{ m s}^{-2}\)
- D \(32.1 \times 10^{11} \text{ m s}^{-2}\)
Answer & Solution
Correct Answer
(B) \(11.5 \times 10^{11} \text{ m s}^{-2}\)
Step-by-step Solution
Detailed explanation
\(F = qvB\) \(a = \frac{F}{m} = \frac{qvB}{m}\) \(a = \frac{(1.6 \times 10^{-19})(4 \times 10^6)(3 \times 10^{-3})}{1.67 \times 10^{-27}}\) \(a = \frac{19.2 \times 10^{-16}}{1.67 \times 10^{-27}}\) \(a \approx 11.5 \times 10^{11} \text{ m s}^{-2}\)
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