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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 \( m = 1.67 \times 10^{-27} \text{ kg} \))

  1. A \( 5.75 \times 10^{11} \text{ m s}^{-2} \)
  2. B \( 5.8 \times 10^{10} \text{ m s}^{-2} \)
  3. C \( 9.6 \times 10^2 \text{ m s}^{-2} \)
  4. D \( 5.8 \times 10^{-11} \text{ m s}^{-2} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( 5.75 \times 10^{11} \text{ m s}^{-2} \)

Step-by-step Solution

Detailed explanation

\(a = \frac{qvB\sin\theta}{m}\) \(a = \frac{(1.6 \times 10^{-19} \text{ C})(4 \times 10^6 \text{ m s}^{-1})(1.5 \times 10^{-3} \text{ T})\sin(90^\circ)}{1.67 \times 10^{-27} \text{ kg}}\) \(a = 5.75 \times 10^{11} \text{ m s}^{-2}\)