Motion of Charged Particle in a Magnetic Field

Click For Summary
SUMMARY

The trajectory of a charged particle moving in a magnetic field is circular when the velocity is in the i direction (v=vi) and the magnetic field is in the -k direction (B=-kB). This conclusion is derived from the Lorentz force equation, F = qv x B, which indicates that the force acting on the particle is always perpendicular to its velocity and the magnetic field. By applying Newton's Second Law, the relationship between the centripetal force and the radius of motion is established as r = mv/(qB), demonstrating that the radius is directly proportional to velocity and inversely proportional to magnetic field strength.

PREREQUISITES
  • Understanding of Newton's Second Law
  • Familiarity with the Lorentz force equation
  • Knowledge of vector cross products
  • Basic concepts of circular motion and centripetal force
NEXT STEPS
  • Study the implications of the Lorentz force in different magnetic field orientations
  • Explore the effects of varying charge and mass on circular motion in magnetic fields
  • Learn about applications of charged particle motion in devices like cyclotrons
  • Investigate the role of magnetic fields in plasma physics
USEFUL FOR

Physics students, educators, and professionals in fields such as electromagnetism, engineering, and applied physics who are interested in the dynamics of charged particles in magnetic fields.

SoberSteve2121
Messages
5
Reaction score
0
If the velocity is in the i direction (v=vi) and the magnetic field is in the -k direction (B=-kB), prove using Newtons Second Law that the trajectory of this charge must be circular? Please help this is really important.
 
Physics news on Phys.org
F = qv X B (X = cross product)

F = ma

So,

qv X B = ma

Solve that and think about your solution...don't lose track of your unit vectors...and think about what they mean, what they tell you...
 
Last edited:


According to Newton's Second Law, the net force acting on a charged particle is equal to the product of its charge and its acceleration. In the case of a charged particle moving in a magnetic field, this net force is given by the Lorentz force equation:

F = qv x B

Where v is the velocity of the particle, B is the magnetic field, and q is the charge of the particle.

Since the velocity of the particle is in the i direction (v=vi) and the magnetic field is in the -k direction (B=-kB), the cross product of v and B can be written as:

v x B = (vi) x (-kB) = -kv x B

This means that the Lorentz force acting on the particle is always perpendicular to both the velocity and the magnetic field. In other words, the particle experiences a centripetal force towards the center of a circle.

According to Newton's Second Law, this centripetal force must be equal to the product of the mass of the particle and its centripetal acceleration:

F = ma = m(v^2/r)

Where m is the mass of the particle, v is its velocity, and r is the radius of the circular motion.

Substituting the Lorentz force equation into this equation, we get:

q(vi) x (-kB) = m(v^2/r)

Simplifying this equation, we get:

qBvi = mv^2/r

Rearranging this equation, we get:

r = mv/(qB)

This shows that the radius of the circular motion is directly proportional to the velocity of the particle and inversely proportional to the magnetic field strength. Therefore, for a given velocity and magnetic field, the trajectory of the charged particle must be circular. This proves that the motion of a charged particle in a magnetic field with a velocity in the i direction and a magnetic field in the -k direction must be circular.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
14
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K