JEE Main201912 Apr 2019Evening ShiftPhysicsMagnetic Effects of CurrentActual
An electron, moving along the x - axis with an initial energy of 100 e V , enters a region of magnetic field B → = 1.5 × 10 - 3 T k ^ at S (see figure). The field extends between x = 0 and x = 2 c m . The electron is detected at the point Q on a screen placed 8 c m away from the point S . The distance d between P and Q (on the screen) is: (electron’s charge 1.6 × 10 - 19 C , mass of electron = 9
Options
- A1.22 c m
- B12.87 c m
- C11.65 c m
- D2.25 c m
Correct answer
B. 12.87 c m
Step-by-step solution
R = m V q B = 2 m ( K . E ) q B m V = 2 m   K . E = 2 × 9.1 × 10 - 31 × 100 × 1.6 × 10 - 19 1.6 × 10 - 19 × 1.5 × 10 - 3 = 2.248   c m ≈   2.25   c m sin ⁡ θ = 2 2.248 = 0.89 P A = R 1 - cos ⁡ θ = 1.22   c m A B = 6 tan ⁡ θ = 11.71   c m d = P A + A B = 12.93   c m