Question

16. Derive an expression for the velocity vC of a positive ions passing undeflected through a region where crossed and uniform electric field E and magnetic field B are simultaneously present. Draw and justify the trajectory of identical positive ions whose velocity has a magnitude

Answer 1

E = E j (j cap as in along the y axis)

B = B k (along the z axis)

Force of positive ion due to electric field, F(due to E) = q Ej

Force of positive ion due to magnetic field, F(due to B) = q (vC x B)k

For passing undeflected, F (due to E) = - F (due to B)

which implies, q Ej = - q (vC x B)k

This is only possible if vC = (E/B) i

For trajectory,

Justification:

For positive ions with speed v < vC,

Force due to electric field F' (due to E) = q E = F (due to E)

Force due to magnetic field F' (due to B) = q vB < F (due to B) since v < vC

Now forces are unbalanced, hence ions will experience an acceleration along E                                    

Answer 2

E = E j and B = B k

Force on positive ion due to electric field F = q E (j)

Force due to magnetic field F  = q (vc x B) (k)

For passing undeflected, FE = - FB

qEj = - q (vc x Bk)

This is possible only if q (vc x B)k = qvcBj

or vc = (E/B)i

For trajectory:-

Justification: For positive ions with speed v <vc

Force due to electric field = F’E = qE = FE

due to magnetic field F’B = qvB <FB since v <vc

Now forces are unbalanced, and hence, ion will experience an acceleration along E.

Since initial velocity is perpendicular to E, the trajectory would be parabolic.