## Related questions with answers

Bainbridge's mass spectrometer, separates ions having the same velocity. The ions, after entering through slits $S_1$ and $S_2$, pass through a velocity selector composed of an electric field produced by the charged plates P and $\mathbf{P}^{\prime}$, and a magnetic field $\overrightarrow{\mathbf{B}}$ perpendicular to the electric field and the ion path. Those ions that pass undeviated through the crossed $\overrightarrow{\mathbf{E}}$ and $\overrightarrow{\mathbf{B}}$ fields enter into a region where a second magnetic field $\vec{B}^{\prime}$ exists, and are bent into circular paths. A photographic plate registers their arrival. Show thal $q / m=E / r B B^{\prime}$, where $r$ is the radius of the circular orbit.

Solution

VerifiedIn the half-disk area of the spectrometer, the magnetic force plays the role of a centripetal force. From this, we can write that

$\frac{mv^2}{R}=qvB'.$

From this, we can first divide by the speed and then rearrange to get the mass as divisor, getting

$\frac{v}{R}=\frac{qB'}{m},$

and finally, the required ratio, as

$\frac{q}{m}=\frac{v}{RB'}.$

Among these, what we can't tell by measurements is the speed of the electrons. In order to obtain it, we will have to consider that in the part where there is electric field as well. Since the particles move in a straight line, this means that the magnetic and the electric forces in this region, which point opposite each other, are of equal magnitudes. From this we can write that

$qE=qvB,$

from where, having simplified the charge, we can obtain the speed as

$v=\frac{E}{B}.$

Substituting the speed at the expression for the charge-to-mass ratio, we can find

$\frac{q}{m}=\frac{E/B}{RB'},$

or finally

$\boxed{\frac{q}{m}=\frac{E}{RBB'}}.$

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