Q1.An electron (mass 9 x 10⁻³¹ kg and charge 1.6 x 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 x 10⁻⁴ T perpendicular to its direction of motion. We wish to apply an uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 x 10⁸ m s⁻¹)
- AE is perpendicular to B and its magnitude is 27 x 10⁴ V m⁻¹
- BE is perpendicular to B and its magnitude is 27 x 10² V m⁻¹Answer
- CE is parallel to B and its magnitude is 27 x 10² V m⁻¹
- DE is parallel to B and its magnitude is 27 x 10⁴ V m⁻¹
This is a velocity selector: the electron passes undeflected only when the electric force qE exactly cancels the magnetic force qv x B, which forces E to be perpendicular to both v and B. Equating magnitudes, E = vB = (3 x 10⁶)(9 x 10⁻⁴) = 27 x 10² V m⁻¹.
Source: NEET 2025 (4 May), Code 45, Q3