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A magnetic field of `(4.0xx10^-3hatk)T` exerts a force `(4.0hati+3.0hatj)xx10^-10N` on a particle having a charge `10^-9C` and moving in te `x-y` plane. Find the velocity of the particle. |
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Answer» Given `B=(4xx10^-3hatk)T,q=10^-9C` and magnetic force `F_m=(4.0hati+3.0hatj)xx10^-10N` Let velocity of the particle in `x-y` plane be `v=v_xhati+v_yhatj` Then, from the relation `F_m=q(vxxB)` We have `(4.0hati+3.0hatj)xx10^-10=10^-9[(v_xhati+v_yhatj)xx(4xx10^-3hatk)]` `=(4v_yxx10^-12hati-4v_x xx 10^-12hatj)` Comparing the coefficients of `hati` and `hatj` we have `4xx10^-10=4v_yxx10^-12` `:. v_y=10^2m//s=100m//s` and `3.0xx10^-10=-4v_x xx10^-12` `:. v_x=-75m//s` `:. v=(-75hati+100hatj)m//s` |
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