1. Tardigrade
  2. Question
  3. Physics
  4. In a certain mass spectrometer, an ion beam passes through a velocity filter consisting of mutually perpendicular fields E and B . The beam then enters a region of another magnetic field B' perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. In a certain mass spectrometer, an ion beam passes through a velocity filter consisting of mutually perpendicular fields $ E $ and $ B $ . The beam then enters a region of another magnetic field $ B' $ perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to

AMUAMU 2018Moving Charges and Magnetism

Solution:

According to the question
In 1st case, velocity of the beam,
$V=\frac{E}{B} \dots(i)$
As we know that, the radius of curvature of a charged particle in a magnetic field
$r=\frac{mv}{qB}$
where, $m$ = mass of the charged particle,
$q$ = charge on the particle,
$B$ = magnitude of the magnetic field
and $v$ = velocity of the particle
In IInd case, $r=\frac{mv}{qB'}=\frac{mE/ B}{q B'}$ [from Eq. $\left(i\right)$]
$=\frac{mE}{qBB'}$
Thus, $r \propto\frac{E}{BB'}$