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) \[\vec{v}=\frac{\vec{E}\times \vec{B}}{{{E}^{2}}}\]
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.
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