CUET · PHYSICS · PYQ PAPER 2025
An electron moves with a velocity of \( 1200 \, \text{m s}^{-1} \) in a magnetic field of 0.4 T at an angle of 30\(^\circ\). What will be the approximate radius of its path? (Take \( e/m \) of electron \( = 1.76 \times 10^{11} \, \text{C kg}^{-1} \))
- A \( 9 \times 10^{-8} \, \text{m} \)
- B \( 8.6 \times 10^{-9} \, \text{m} \)
- C \( 4.4 \times 10^{-10} \, \text{m} \)
- D \( 1.5 \times 10^{-11} \, \text{m} \)
Answer & Solution
Correct Answer
(B) \( 8.6 \times 10^{-9} \, \text{m} \)
Step-by-step Solution
Detailed explanation
\( r = \frac{v \sin\theta}{(e/m) B} \) \( r = \frac{1200 \, \text{m s}^{-1} \times \sin(30^\circ)}{(1.76 \times 10^{11} \, \text{C kg}^{-1}) \times 0.4 \, \text{T}} \) \( r = \frac{1200 \times 0.5}{1.76 \times 10^{11} \times 0.4} \, \text{m} \)…
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