InterviewSolution
Saved Bookmarks
| 1. |
Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m` A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces. y-coordinate of particle where it collides with screen (in meters) isA. `(0.6(sqrt3-1))/(sqrt3)`B. `(0.6(sqrt3+1))/(sqrt3)`C. `1.2(sqrt3+1)D. `(1.2(sqrt3-1))/(sqrt3)` |
|
Answer» Correct Answer - d |
|