In J.J. Thomson’s method, electric field $E$ , magnetic field $B$ and velocity $v$ of the electron were in mutually perpendicular direction. This velocity selector allows particles of velocity $v$ to pass undeflected when
(correct answer + 1, wrong answer - 0.25)

v = B E

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v = \frac{E}{B}

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v = \frac{B}{E}

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v = \frac{B^{2}}{E}

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Solution

The correct option is B $v = \frac{E}{B}$
Electric force $= E e$
Magnetic force $= B e v$
If both of them act in opposite directions, they may cancel each other. Then
$E e = B v e$
$v = \frac{E}{B}$

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In J.J. Thomson’s method, electric field E, magnetic field B and velocity v of the electron were in mutually perpendicular direction. This velocity selector allows particles of velocity v to pass undeflected when (correct answer + 1, wrong answer - 0.25)

The correct option is B v=EBElectric force =Ee Magnetic force =Bev If both of them act in opposite directions, they may cancel each other. Then Ee=Bve v=EB

In J.J. Thomson’s method, electric field $E$ , magnetic field $B$ and velocity $v$ of the electron were in mutually perpendicular direction. This velocity selector allows particles of velocity $v$ to pass undeflected when
(correct answer + 1, wrong answer - 0.25)

The correct option is B $v = \frac{E}{B}$
Electric force $= E e$
Magnetic force $= B e v$
If both of them act in opposite directions, they may cancel each other. Then
$E e = B v e$
$v = \frac{E}{B}$