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In Thomson mass spectrograph $\vec{E} \perp \vec{B}$ then the velocity of undeflected electron beam will be :
A
$\dfrac{|\vec{E}|}{|\vec{B}|}$
B
$\vec{E} \times \vec{B}$
C
$\dfrac{|\vec{B}|}{|\vec{E}|}$
D
$\dfrac{E^2}{B^2}$
Detailed Solution
In crossed electric and magnetic fields, the electric force eE and the magnetic force evB on the electron act in opposite directions.
The beam goes undeflected when these forces balance: $eE = evB$
$\Rightarrow v = \dfrac{E}{B}$
So the velocity of the undeflected beam is $\dfrac{|\vec{E}|}{|\vec{B}|}$.
The beam goes undeflected when these forces balance: $eE = evB$
$\Rightarrow v = \dfrac{E}{B}$
So the velocity of the undeflected beam is $\dfrac{|\vec{E}|}{|\vec{B}|}$.