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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. Time after which the particle will collide the screen is (in seconds)A. `1/5 (3+(pi)/(6)+(1)/(sqrt(3)))`B. `1/5 (6+(pi)/(3)+(sqrt(3)))`C. `1/3 (5+(pi)/(6)+(1)/(sqrt(3)))`D. `1/5 (6+(pi)/(18)+(sqrt(3)))` |
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Answer» Correct Answer - A `t_(OA)=sqrt((2s)/(a))=sqrt((2sm)/(qE))=sqrt((2 xx 1.8 xx 1)/((1)(10)))` `=0.6 sec=3//5 sec` `t_(AB)=(30^(@))/(360^(@))T= 1/12 xx (2pi m)/(qB)` `((2pi)(1))/((12)(1)(5))=(pi)/(30)sec` `t_(BC)=(BC)/(v)=(0.6 sec theta)/(v)=(0.6((2)/(sqrt(3)))/(6))=(1)/(5sqrt(3)) sec` Hence `3/5+(pi)/(30)+(1)/(5sqrt(3))=(1)/(5)(3+(pi)/(6)+(1)/(sqrt(3))) sec` Hence choice (a) is correct. |
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