A particle of mass m and charge q is projected from the origin with a velocity v0j^ in a region where an electric field is given by E→=E0i^, at time t=0. Then choose the correct statement(s). (Given mv02=2qE0x0).

u=v0j^ E=E0i^ a→=9E0mi^Radius of curvature R=v2a⊥velocity along x-axis, when x=x0ux=2ax0=2aex0m=v0 vy=v0Net speed= vx2+vy2=v02 R=(v02)2a cos45°=m(v02)229e0=v0242m9e0 29e0x09e042=42x0

A particle of mass m and charge q is projected from the origin with a velocity v0j^ in a region where an electric field is given by E→=E0i^, at time t=0. Then choose the correct statement(s). (Given mv02=2qE0x0).

u=v0j^ E=E0i^ a→=9E0mi^Radius of curvature R=v2a⊥velocity along x-axis, when x=x0ux=2ax0=2aex0m=v0 vy=v0Net speed= vx2+vy2=v02 R=(v02)2a cos45°=m(v02)229e0=v0242m9e0 29e0x09e042=42x0

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A particle of mass $m$ and charge $q$ is projected from the origin with a velocity $v_{0} \hat{j}$ in a region where an electric field is given by $\vec{E} = E_{0} \hat{i}$ , at time $t = 0$ . Then choose the correct statement(s). (Given $m {v_{0}}^{2} = 2 q E_{0} x_{0}$ ).

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Radius of curvature of the particle when its $x$ coordinate becomes $x_{0}$ is $2 x_{0 .}$

Radius of curvature of the particle when its $x$ coordinate becomes $x_{0}$ is $4 \sqrt{2} x_{0}$ .

Speed of the particle when its $x$ coordinate becomes $x_{0}$ is $\sqrt{2} v_{0}$ .

Speed of the particle when its $x$ coordinate becomes $x_{0}$ is $2 v_{0 .}$

answer is B, C.

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Detailed Solution

$u = v_{0} \hat{j} E = E_{0} \hat{i} \vec{a} = \frac{9 E_{0}}{m} \hat{i}$

Radius of curvature $R = \frac{v^{2}}{a_{⊥}}$

velocity along $x$ -axis, when $x = x_{0}$

$u_{x} = \sqrt{2 a x_{0}} = \sqrt{\frac{2 a e x_{0}}{m}} = v_{0} v_{y} = v_{0}$

Net speed= $\sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = v_{0} \sqrt{2} R = \frac{{(v_{0} \sqrt{2})}^{2}}{a \cos 45 °} = \frac{m {(v_{0} \sqrt{2})}^{2} \sqrt{2}}{9 e_{0}} = {v_{0}}^{2} 4 \sqrt{2} \frac{m}{9 e_{0}} \frac{29 e_{0} x_{0}}{9 e_{0}} 4 \sqrt{2} = 4 \sqrt{2} x_{0}$