Magnetic Effects Of Current Question 414
Question: A charged particle of specific charge (charge/ mass) $ \alpha $ is released from origin at time t = 0 with velocity $ \overset{\to }{\mathop{v}},=v_{0}(\hat{i}+\hat{j}) $ in uniform magnetic field $ \overset{\to }{\mathop{B}},=B_{0}\hat{i} $ . Coordinates of the particle at time $ t=\pi /(B_{0}\alpha ) $
Options:
A) $ ( \frac{v_{0}}{2B_{0}\alpha },\frac{\sqrt{2}v_{0}}{\alpha B_{0}},\frac{-v_{0}}{B_{0}\alpha } ) $
B) $ ( \frac{-v_{0}}{2B_{0}\alpha },0,0 ) $
C) $ ( 0,\frac{2v_{0}}{B_{0}\alpha },\frac{v_{0}\pi }{2B_{0}\alpha } ) $
D) $ ( \frac{v_{0}\pi }{B_{0}\pi },0\frac{-2v_{0}}{B_{0}\alpha } ) $
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Answer:
Correct Answer: D
Solution:
[d] $ \alpha =\frac{q}{m} $ , path of the particle will be a helix of time period, $ T=\frac{2\pi m}{B_{0}q}=\frac{2\pi }{B_{0}\alpha } $ The give time $ t=\frac{\pi }{B_{0}\alpha }=\frac{T}{2} $
$ \therefore $ Coordinates of particle at time $ t=T/2 $ would be $ (vx,T/2,,0,-2r) $ Here, $ r=\frac{mv_{0}}{B_{0}q}=\frac{v_{0}}{B_{0}\alpha } $
$ \therefore $ The coordinate are $ ( \frac{v_{0}\pi }{B_{0}\alpha },0,\frac{-2v_{0}}{B_{0}\alpha } ) $
BETA
