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Question

Physics Question on Dual nature of matter

In Thomson mass spectrograph E⃗⊥B⃗\vec{E}\perp \vec{B} then the velocity of electron beam will be

A

∣E‾∣∣B‾∣\frac {|\overline{E}|}{|\overline{B}|}

B

E‾×B‾\overline{E}\times \overline{B}

C

∣B‾∣∣E‾∣\frac {|\overline{B}|}{|\overline{E}|}

D

E2‾B2‾\frac {\overline{E^2}}{\overline{B^2}}

Answer

∣E‾∣∣B‾∣\frac {|\overline{E}|}{|\overline{B}|}

Explanation

Solution

eE = evB ∴v=∣E∣∣B∣\therefore v=\frac {|E|}{|B|}