Magnetic Effects Of Current Question 240
Question: A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields $ \vec{E} $ and $ \vec{B} $ with a velocity $ \vec{v} $ perpendicular to both $ \vec{E} $ and $ \vec{B} $ , and comes out without any change in magnitude or direction of $ \vec{v} $ . Then
Options:
A) $ \vec{v}=\frac{\vec{B}\times \vec{E}}{E^{2}} $
B) $ \vec{v}=\frac{\vec{E}\times \vec{B}}{B^{2}} $
C) $ \vec{v}=\frac{\vec{B}\times \vec{E}}{B^{2}} $
D) $ \vec{v}=\frac{\vec{E}\times \vec{B}}{E^{2}} $
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Answer:
Correct Answer: B
Solution:
[b] When $ \vec{E} $ and $ \vec{B} $ are perpendicular and velocity has no changes then $ qE=qvB $ i.e., $ v=\frac{E}{B}. $ The two forces oppose each other if v is along $ \vec{E}\times \vec{B} $ i.e., $ \vec{v}=\frac{\vec{E}\times \vec{B}}{B^{2}} $ as $ \vec{E} $ and $ \vec{B} $ are perpendicular to each other. $ \frac{\vec{E}\times \vec{B}}{B^{2}}=\frac{EB\sin 90{}^\circ }{B^{2}}=\frac{E}{B} $ For historic and standard experiments like Thomson?s e/m value, if v is given only as E/B, it would have been better form the pedagogic view, although the answer is numerically correct.
BETA
