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Physics · PYQ NEET UG 2025

हिन्दी

An electron (mass \( 9 \times 10^{-31}\text{ kg} \) and charge \( 1.6 \times 10^{-19}\text{ C} \)) moving with speed \( c/100 \) (\( c = \) speed of light) is injected into a magnetic field \( \vec{B} \) of magnitude \( 9 \times 10^{-4}\text{ T} \) perpendicular to its direction of motion. We wish to apply an uniform electric field \( \vec{E} \) together with the magnetic field so that the electron does not deflect from its path. Then (speed of light \( c = 3 \times 10^8\text{ ms}^{-1} \))

  1. \( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 27 \times 10^4\text{ V m}^{-1} \)
  2. \( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 27 \times 10^2\text{ V m}^{-1} \)
  3. \( \vec{E} \) is parallel to \( \vec{B} \) and its magnitude is \( 27 \times 10^2\text{ V m}^{-1} \)
  4. \( \vec{E} \) is parallel to \( \vec{B} \) and its magnitude is \( 27 \times 10^4\text{ V m}^{-1} \)

Correct answer · Option 2

\( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 27 \times 10^2\text{ V m}^{-1} \)

Explanation

For no deflection, \( E = vB = (3 \times 10^6 / 100) \times 9 \times 10^{-4} = 2.7 \times 10^3\text{ V/m} = 27 \times 10^2\text{ V m}^{-1} \) and \( \vec{E} \) must be perpendicular to both \( \vec{v} \) and \( \vec{B} \).
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