In Thomson mass spectrograph \overrightarrow{E} \perp \overrightarrow{B} then the velocity of electron beam will be (2001)

Correct Answer is : |→E||→B|

\overrightarrow{| {E} |}/\overrightarrow{| {B} |}

{E} ^{\to} \times\overrightarrow{B}

\overrightarrow{| {B} |}/\overrightarrow{| {E} |}

\overrightarrow{ {E} ^{2}}/\overrightarrow{ {B} ^{2}}

Class 12 NEET Physics Dual Nature of Matter and Radiation

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In Thomson mass spectrograph

→

⊥

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then the velocity of electron beam will be

(2001)

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