A) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{E}^{2}}}\]
B) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\]
C) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{E}}{{{E}^{2}}}\]
D) \[\overrightarrow{v}=\frac{\overrightarrow{B}\times \overrightarrow{E}}{{{B}^{2}}}\]
Correct Answer: B
Solution :
(b) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\] Force acting on a particle under both electric and magnetic field is \[\overrightarrow{F}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})=0\](since neither direction nor magnitude changes) Taking cross product with \[\overrightarrow{B}\] on both sides. \[0=q\overrightarrow{E}\times \overrightarrow{B}+q(\overrightarrow{v}\times \overrightarrow{B})\times \overrightarrow{B}\] \[=q\overrightarrow{E}\times \overrightarrow{B}+q(\overrightarrow{v}.\overrightarrow{B})\overrightarrow{B}-\overrightarrow{v}\,{{B}^{2}}\] \[\overrightarrow{v}=\overrightarrow{E}\times \overrightarrow{B}/{{B}^{2}}(\because \overrightarrow{v}\,.\,\overrightarrow{B}=0)\]You need to login to perform this action.
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