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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. The speed with which the particle will collide the screen isA. `3 m s^(-1)`B. `6 m// s^(-1)`C. `9 m//s^(-1)D. `12 m// s^(-1)` |
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Answer» Correct Answer - B Velocity at `A v=sqrt(2 as)` `=sqrt(2xx((qE)/(m))s) = sqrt((2 xx 1 xx 10 xx 1.8)/(1))=6 m//s` In magnetic field, speed does not change. Hence particle will collide with speed `6m//s`. |
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