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

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 (A)  E B'/B  (B) E B/B'  (C)  BB'/E  (D)  E/BB'

Tardigrade Admin · Accepted Answer

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{m v}{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{m v}{q B ^{'}} = \frac{m E / B}{q B ^{'}}

[from Eq.

(i)

]

= \frac{m E}{qB B ^{'}}

Thus,

r \propto \frac{E}{B B ^{'}}

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

$E B^{'} / B$

$EB / B^{'}$

$B B^{'} / E$

$E / B B^{'}$

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

EB′/B

EB/B′

BB′/E

E/BB′

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

$E B^{'} / B$

$EB / B^{'}$

$B B^{'} / E$

$E / B B^{'}$