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Question
In Thomson mass spectrograph $\overrightarrow E \bot \overrightarrow B $ then the velocity of electron beam will be
${{\left| {\overrightarrow E } \right|} \over {\left| {\overrightarrow B } \right|}}$
$\overrightarrow E \times \overrightarrow B $
${{\left| {\overrightarrow B } \right|} \over {\left| {\overrightarrow E } \right|}}$
${{{{\overrightarrow E }^2}} \over {{{\overrightarrow B }^2}}}$

Solution

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