CUET · PHYSICS · PYQ PAPER 2025
A proton is projected with a speed of \( v = 4 \times 10^6 \text{ m s}^{-1} \) horizontally from east to west. A uniform magnetic field \( \vec{B} \) of strength \( B = 1.5 \times 10^{-3} \text{ T} \) exists in the vertically upward direction. What is the acceleration produced? (Use charge on the proton \( q = 1.6 \times 10^{-19} \text{ C} \) and mass of the proton \( m = 1.67 \times 10^{-27} \text{ kg} \))
- A \( 5.75 \times 10^{11} \text{ m s}^{-2} \)
- B \( 5.8 \times 10^{10} \text{ m s}^{-2} \)
- C \( 9.6 \times 10^2 \text{ m s}^{-2} \)
- D \( 5.8 \times 10^{-11} \text{ m s}^{-2} \)
Answer & Solution
Correct Answer
(A) \( 5.75 \times 10^{11} \text{ m s}^{-2} \)
Step-by-step Solution
Detailed explanation
\(a = \frac{qvB\sin\theta}{m}\) \(a = \frac{(1.6 \times 10^{-19} \text{ C})(4 \times 10^6 \text{ m s}^{-1})(1.5 \times 10^{-3} \text{ T})\sin(90^\circ)}{1.67 \times 10^{-27} \text{ kg}}\) \(a = 5.75 \times 10^{11} \text{ m s}^{-2}\)
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