A) \[\vec{v}=\vec{E}\times \vec{B}/{{B}^{2}}\]
B) \[\vec{v}=\vec{B}\times \vec{E}/{{B}^{2}}\]
C) \[\vec{v}=\vec{E}\times \vec{B}/{{E}^{2}}\]
D) \[\vec{v}=\vec{B}\times \vec{E}/{{E}^{2}}\]
Correct Answer: A
Solution :
As v of charged particle is remaining constant, it means, force acting on charged particle is zero. So, \[{{\tan }^{-1}}\frac{bc}{a(c-a)}\] \[{{\tan }^{-1}}\frac{bc}{a}\]\[{{y}^{2}}=8x\] \[y=x+2\]\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-12y-2\text{ }z+20=0,\]You need to login to perform this action.
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