CBSE Class 12 Physics 2019 Delhi Set 3 Paper

Question : 4 of 8

Marks: +1, -0

A charged particle

q

is moving in the presence of a magnetic field

B

which is inclined to an angle

30^{∘}

with the direction of motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path.

Solution:

Since magnetic force is given as,

\;{F}↖{→}\;=q({v}↖{→} × {B}↖{→})

\;\text " or "\;\;F\;=q v B sin \;θ

\;\text " here, "\;\;θ\;=30^{∘}

\;F\;=\;{q v B}/{2}\

Velocity component parallel to the field moves particle in straight line and velocity component perpendicular to the field creates circular motion thus the combined effect is the helical path.

Go to Question:

Prev Question

Next Question