Question

In J.J.Thomson’s method, electric field $E$ , magnetic field $B$ and velocity $v$ of the electrons were in mutually perpendicular direction. The velocity selector allows particles of velocity $v$ to pass undeflected when
(A) $v = BE$
(B) $v = \dfrac{E}{B}$
(C) $v = \dfrac{B}{E}$
(D) $v = \dfrac{{{B^2}}}{E}$

Answer 1

Hint: We are asked to find the velocity of the velocity selector. Thus we will equate the electric force and magnetic force and then we will come up with a relation.
Formula Used
${F_E} = Ee$
Where, ${F_E}$ is the electric force, $E$ is the electric field and $e$ is the charge on an electron or proton.
${F_M} = Bve$
Where, ${F_B}$ is the magnetic force, $B$ is the magnetic field and $v$ is the velocity of the charge.

Step By Step Solution
We know,
The electric force on a charge, ${F_E} = \dfrac{{k{e^2}}}{{{r^2}}}$
The electric field on a charge, $E = \dfrac{{ke}}{{{r^2}}}$
Thus, we can say
${F_E} = Ee \cdot \cdot \cdot \cdot (1)$
Also, we know
Magnetic force on a charge, ${F_M} = Bve \cdot \cdot \cdot \cdot (2)$
Now,
Equating $(1)$and $(2)$,
${F_E} = {F_M}$

Substituting the value of electric force and magnetic force,
$Ee = Bve$
Cancelling $e$ and evaluating further, we get
$v = \dfrac{E}{B}$

Hence, the answer is (B).

Additional Information
A velocity selector consists of perpendicular magnetic fields. It is also known as Wien’s filter.
As the name suggests, the filter allows charges only with certain velocity and beyond to pass through it. This can help us to overthrow the charges having lower velocity from the velocity we calculated. We can construct a filter according to our will as what minimum velocity of charge do we want to flow through.

Note: In this case, we were only asked to find out the threshold velocity of the selector. If we were asked to find out something other than the threshold velocity, then the equal sign becomes a sign of comparison.