Magnetic Effects Of Current Question 409
Question: A particle of charge q and mass m starts moving from the origin under the action of an electric field $ \overset{\to }{\mathop{E}},=E_{0}\hat{i} $ and $ \overset{\to }{\mathop{B}},=B_{0}\hat{i} $ with velocity $ \overset{\to }{\mathop{v}},=v_{0}\hat{j} $ .The speed of the particle will become $ 2v_{0} $ after time
Options:
A) $ t=\frac{2mv_{0}}{qE} $
B) $ t=\frac{2Bq}{mv_{0}} $
C) $ t=\frac{\sqrt{3}Bq}{mv_{0}} $
D) $ t=\frac{\sqrt{3}mv_{0}}{qE} $
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Answer:
Correct Answer: D
Solution:
[d] Electric force on the particle, $ F=Eq $ , and displacement $ s=\frac{1}{2}at^{2}=\frac{1}{2}( \frac{Eq}{m} )t^{2} $ . Now, $ W=\Delta K $ , or $ Fs=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2}) $ or $ Eq\times \frac{1}{2}( \frac{Eq}{m} )t^{2} $ $ =\frac{1}{2}m[{{(2v_{0})}^{2}}-v_{0}^{2}] $
$ \therefore t=\frac{\sqrt{3}mv_{0}}{qE} $
BETA
