Magnetic Effects Of Current Question 249
Question: An electric field acts along positive x-axis. A charged particle of charge q and mass m is released from origin and moves with velocity $ \vec{v}=v_{0}\hat{j} $ under the action of electric’ field and magnetic field, $ \vec{B}=B_{0}\hat{i} $ . The velocity of particle becomes 2 $ v_{0} $ after time $ \frac{\sqrt{3}mv_{0}}{\sqrt{2}qE_{0}} $ . Find the electric field.
Options:
A) $ \frac{\sqrt{2}}{\sqrt{3}}E_{0}\hat{i} $
B) $ \frac{\sqrt{3}}{\sqrt{2}}E_{0}\hat{i} $
C) $ \sqrt{3}E_{0}\hat{i} $
D) $ \sqrt{2}E_{0}\hat{i} $
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Answer:
Correct Answer: D
Solution:
[d] Path of particle is helix with increasing pitch $ v={{(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}^{1/2}} $ Here $ v_{x}^{2}={{( \frac{qE}{m}t )}^{2}} $ and $ v_{y}^{2}+v_{z}^{2}=v_{0}^{2} $ Also $ v=2v_{0} $ $ {{(2v_{0})}^{2}}=\frac{q^{2}E^{2}t^{2}}{m^{2}}+v_{0}^{2} $ $ t=\frac{\sqrt{3}mv_{0}}{qE}=\frac{\sqrt{3}mv_{0}}{\sqrt{2}qE_{0}}given $ $ \vec{E}=\sqrt{2}E_{0}\hat{i} $
BETA
