The plane RS divides space into a region where a uniform electric field of intensity E exists (in the plane of figure and perpendicular to RS and above RS), and a region where a uniform magnetic field of induction B acts (perpendicular to plane of figure as shown and below RS). A positively charged particle is launched with a speed v from the plane towards the electric field at an angle α to the plane RS as shown. At what angle values α in degrees (0

Question

The plane RS divides space into a region where a uniform electric field of intensity E exists (in the plane of figure and perpendicular to RS and above RS), and a region where a uniform magnetic field of induction B acts (perpendicular to plane of figure as shown and below RS). A positively charged particle is launched with a speed v from the plane towards the electric field at an angle α  to the plane RS as shown. At what angle values α  in degrees (0<α<π)  will the particle trajectory represent a closed path? It is given that  EBv=12.

Accepted Answer

When a charged particle moves in a region of a uniform electric field of intensity $E,$ it is acted upon by a constant force $q E$ directed towards the plane $R S .$ The particle moves along a parabola (see Fig.) with constant acceleration,
Question Image

$m a = q E \Leftrightarrow a = \frac{q E}{m} .$

The time it takes to move along a parabola is:

$T_{E} = \frac{2 v \sin α}{a} = \frac{2 m v \sin α}{q E} L_{E} = \frac{v^{2} \sin 2 α}{a} = \frac{m v^{2} \sin 2 α}{q E} L_{B} = 2 R \sin α = \frac{2 m v \sin α}{q B}$

But in the given situation $B$ is in opposite direction for complete path $L E = L B$ Hence

$α = π - \cos^{- 1} (\frac{E}{B v}) = 120$

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