When `Euarr uarrB`e and particle velocity is perpendicuular to both of these field.
Text Solution
AI Generated Solution
To solve the problem, we need to analyze the motion of a charged particle in the presence of both electric and magnetic fields when the velocity of the particle is perpendicular to both fields. Let's break this down step by step.
### Step 1: Understanding the Setup
We have:
- An electric field \( \mathbf{E} \) in the y-direction.
- A magnetic field \( \mathbf{B} \) in the x-direction.
- The velocity \( \mathbf{v} \) of the particle is perpendicular to both \( \mathbf{E} \) and \( \mathbf{B} \).
...
Topper's Solved these Questions
MAGNETICS
DC PANDEY ENGLISH|Exercise INTRODUCTORY EXERCISE|1 Videos
MAGNETICS
DC PANDEY ENGLISH|Exercise SUBJECTIVE_TYPE|1 Videos
MAGNETIC FIELD AND FORCES
DC PANDEY ENGLISH|Exercise Medical entrance s gallery|59 Videos
MAGNETISM AND MATTER
DC PANDEY ENGLISH|Exercise Medical gallery|1 Videos
Similar Questions
Explore conceptually related problems
Assertion When a charged particle moves perpendicular to a uniform magnetic field then its momentum remains constant. Reason Magnetic force acts perpendicular to the velocity of the particle.
An (alpha) -particle and a proton are both simultaneously projected in opposite direction into a region of constant magnetic field perpendicular to the direction of the field. After some time it is found that the velocity of the (alpha) -particle has changed in a direction by 45^(@) . Then at this time, the angle between velocity vectors of (alpha) -particle and proton is
An (alpha) -particle and a proton are both simultaneously projected in opposite direction into a region of constant magnetic field perpendicular to the direction of the field. After some time it is found that the velocity of the (alpha) -particle has changed in a direction by 45^(@) . Then at this time, the angle between velocity vectors of (alpha) -particle and proton is
A charged particle enters into a uniform magnetic field with velocity v_(0) perpendicular to it , the length of magnetic field is x=sqrt(3)/(2)R , where R is the radius of the circular path of the particle in the field .The magnitude of charge in velocity of the particle when it comes out of the field is
A charged particle enters a uniform magnetic field with velocity v_(0) = 4 m//s perpendicular to it, the length of magnetic field is x = ((sqrt(3))/(2)) R , where R is the radius of the circular path of the particle in the field. Find the magnitude of charge in velocity (in m/s) of the particle when it comes out of the field.
When a charged particle moving with velocity vec v is subjected to a magnetic field of induction vec B , the force on it is non-zero. This implies that:
A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendicular to the field direction. The path of the particle will be
Which of the follwing particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?
Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?
Which of the follwing particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?
