Question

Bainbridge’s mass spectrometer, shown in Fig. 28-54, separates ions having the same velocity. The ions, after entering through slits, S₁and S₂, pass through a velocity selector composed of an electric field produced by the charged plates Pand P', and a magnetic field $\overrightarrow{B}$perpendicular to the electric field and the ion path. The ions that then pass undeviated through the crossed$\overrightarrow{E}$and $\overrightarrow{B}$fields enter into a region where a second magnetic field $\overrightarrow{B}$exists, where they are made to follow circular paths. A photographic plate (or a modern detector) registers their arrival. Show that, for the ions, q/m=E/rBB,where ris the radius of the circular orbit.

Short answer

For Bainbridge’s mass spectrometer, $\frac{q}{m}=\frac{E}{\left(rBB\text{'}\right)}$

Step-by-step solution

Given

Bainbridge’s mass spectrometer contains magnetic field B, which is perpendicular to the electric field E and path of ion. There is a magnetic field B’, which makes the ions to follow a circular path.

Understanding the concept

By using the concept of crossed fields (electric and magnetic) and circulating charged particle in and magnetic field, we will prove that$\frac{q}{m}=\frac{E}{\left(rBB\text{'}\right)}$

Formula:

Magnetic force $F_B=qvB$

Electric force$F_E=qE$

Centripetal force$F=\left(m{v}^2\right)/r$

Show that, for the ions, q/m=E/rBB , where r is the radius of the circular orbit.

In Bainbridge’s mass spectrometer, ions are moving with velocity v in the presence of both an electric field E and magnetic field B. Hence ions experience both:

Magnetic force $F_B=qvB$

Electric force $F_E=qE$

Thus the Lorentz force will be

$F=qE+qvB$

If we adjust two fields such that they cancel each other,

F=0

$\begin{array}{l}qE=qvB\\ v=\frac{E}{B}·\dots ·\left(1\right)\end{array}$

Now, after passing through the crossed field, the ions are made to follow a circular path by applying magnetic field B’. Centripetal force is provided by the magnetic field B’.

Magnetic force,

$F_{B\text{'}}=qvB\text{'}$

Centripetal force,

$F=\left(m{v}^2\right)/r$

Equating these two equations, we get

$qv{B}^{\text{'}}=\frac{\left(m{v}^2\right)}{r}$

Rearranging the equation,

$\frac{q}{m}=\frac{v}{(B\text{'}r)}$

Using equation 1, we get

$\frac{q}{m}=\frac{E}{(rBB\text{'})}$

And hence it is proved.