please answer this :

Question

In a certain mass spectrograph, 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) EB'/B

B) EB/B'

C) BB'/E

D) B/EB'

E) E/BB'

Answer 1

hello Dorsa,

in velocity filter only those ions can pass which satisfy the condition

v= E/B

where,

v= velocity of ion , E = Electric field, B = Magnetic field

Next it passes though another magnetic field B'

Then mvB' = mv²/r

Where m = mass of ion, v= velocity of ion, B' new magnetic field, r= radius of trajectory,

this gives B' =v/r

so r = v/B'

Now plug in v=E/B

r= E/BB'