A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a
helix
cycloid
straight line
circle
A
circle
B
straight line
C
helix
D
cycloid
Open in App
Solution
Verified by Toppr
→E∥→B FE=q→E $F_B= q(\vec{v} \times \vec{B}$ given →v is parallel to →B ⇒FB=0 since sinθ=sin0∘=0 ∴ the charged particle will accelerate and move in a straight line.
Was this answer helpful?
0
Similar Questions
Q1
A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a
View Solution
Q2
A charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other. The particle will move in a?
View Solution
Q3
A charged particle is released from the rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a
View Solution
Q4
A charge particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a
View Solution
Q5
In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a