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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

In J.J.Thomso...

Question

In J.J.Thomson's method , electric field E, magnetic field B and velocity v of the electrons were in mutually perpendicular direction. This velocity selector allows particles of velocity v to pass undeflected when

v = B E

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v = E / B

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v = B / E

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v = B^{2} / E

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Solution

The correct option is C $v = E / B$
$E l e c t r i c F o r c e = E e$
$m a g n e t i c F o r c e = B v e$
If both of them act in opposite directions, they may cancel each other in magnitude
$E e = B v e$
$v = \frac{E}{B}$

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In J.J.Thomson's method , electric field E, magnetic field B and velocity v of the electrons were in mutually perpendicular direction. This velocity selector allows particles of velocity v to pass undeflected when

The correct option is C v=E/BElectricForce=EemagneticForce=BveIf both of them act in opposite directions, they may cancel each other in magnitudeEe=Bvev=EB

The correct option is C $v = E / B$
$E l e c t r i c F o r c e = E e$
$m a g n e t i c F o r c e = B v e$
If both of them act in opposite directions, they may cancel each other in magnitude
$E e = B v e$
$v = \frac{E}{B}$