A +Ve charge ‘q’ with mass ‘m’ is released from origin in a coordinate system with a uniform electric field E→=E0j^ and magnetic field B→=B0k^ exist. Find the displacement of particle along y-axis when its velocity becomes ⊥r to electric field

ax=qϑB0sinθm=qϑyB0m ay=qE0−qϑB0cosθm=qE0−qϑxB0m ϑx=[qE0m−dvydt]mqB0 ax=qE0mVy

A +Ve charge ‘q’ with mass ‘m’ is released from origin in a coordinate system with a uniform electric field E→=E0j^ and magnetic field B→=B0k^ exist. Find the displacement of particle along y-axis when its velocity becomes ⊥r to electric field

ax=qϑB0sinθm=qϑyB0m ay=qE0−qϑB0cosθm=qE0−qϑxB0m ϑx=[qE0m−dvydt]mqB0 ax=qE0mVy

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A +Ve charge ‘q’ with mass ‘m’ is released from origin in a coordinate system with a uniform electric field $\vec{E} = E_{0} \hat{j}$ and magnetic field $\vec{B} = B_{0} \hat{k}$ exist. Find the displacement of particle along y-axis when its velocity becomes $⊥ r$ to electric field

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$y = \frac{m E_{0}}{2 q B_{0}^{2}}$

$y = \frac{2 m E_{0}}{q B_{0}^{2}}$

$y = \frac{m E_{0}}{4 q B_{0}^{2}}$

$y = \frac{m E_{0}}{q B_{0}^{2}}$

answer is A.

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Detailed Solution

$a_{x} = \frac{q ϑ B_{0} \sin θ}{m} = \frac{q ϑ_{y} B_{0}}{m} a_{y} = \frac{q E_{0} - q ϑ B_{0} \cos θ}{m} = \frac{q E_{0} - q ϑ_{x} B_{0}}{m} ϑ_{x} = [\frac{q E_{0}}{m} - \frac{d v_{y}}{d t}] \frac{m}{q B_{0}} a_{x} = \frac{q E_{0}}{m} V_{y}$