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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?

  1. A 14 cm
  2. B 28 cm
  3. C 56 cm
  4. D 112 cm
Verified Solution

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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