Question
A particle of charge $$q$$ and mass $$m$$ starts moving from the origin under the action of an electric field $$\overrightarrow E = {E_0}\hat i$$ and $$\overrightarrow B = {B_0}\hat i$$ with velocity $$\overrightarrow v = {v_0}\hat j.$$ The speed of the particle will become $$2{v_0}$$ after a time
A.
$$t = \frac{{2m{v_0}}}{{qE}}$$
B.
$$t = \frac{{2Bq}}{{m{v_0}}}$$
C.
$$t = \frac{{\sqrt 3 Bq}}{{m{v_0}}}$$
D.
$$t = \frac{{\sqrt 3 m{v_0}}}{{qE}}$$
Answer :
Solution :
Electric force on the particle, $$F = Eq,$$ and displacement $$s = \frac{1}{2}a{t^2} = \frac{1}{2}\left( {\frac{{Eq}}{m}} \right){t^2}.$$
Now, $$W = \Delta K,$$
$$\eqalign{
& {\text{or}}\,\,Fs = \frac{1}{2}m\left( {v_f^2 - v_i^2} \right)\,\,{\text{or}}\,\,Eq \times \frac{1}{2}\left( {\frac{{Eq}}{m}} \right){t^2} \cr
& = \frac{1}{2}m\left[ {{{\left( {2{v_0}} \right)}^2} - v_0^2} \right] \cr
& \therefore t = \frac{{\sqrt 3 m{v_0}}}{{qE}}. \cr} $$