91Ó°ÊÓ

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 µþ’, which makes the ions to follow a circular path.

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

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

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

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

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.