Motion of Charged Particle in a Magnetic Field

In summary, when a charged particle enters a magnetic field, it will experience a force perpendicular to both the field and its motion, resulting in a circular or helical path. The radius of the path is directly proportional to the particle's velocity, described by the equation r = mv/qB. The strength of the magnetic field affects the motion of the particle, with a stronger field resulting in a smaller radius and higher velocity. A charged particle can change its direction of motion in a magnetic field, but cannot experience a net force.
  • #1
SoberSteve2121
5
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
  • #2
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:
  • #3


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.
 

1. How does a charged particle behave in a magnetic field?

When a charged particle enters a magnetic field, it will experience a force perpendicular to both the magnetic field and the direction of its motion. This force causes the particle to move in a circular or helical path depending on the strength and orientation of the magnetic field.

2. What is the relationship between the velocity of a charged particle and the radius of its path in a magnetic field?

The radius of a charged particle's path in a magnetic field is directly proportional to its velocity. This means that as the velocity increases, the radius of the path also increases. This relationship is described by the equation r = mv/qB, where r is the radius, m is the mass of the particle, v is its velocity, q is its charge, and B is the strength of the magnetic field.

3. Can a charged particle change its direction of motion in a magnetic field?

Yes, a charged particle can change its direction of motion in a magnetic field due to the force it experiences. This change in direction is always perpendicular to both the magnetic field and the particle's original velocity, resulting in a circular or helical path.

4. How does the strength of the magnetic field affect the motion of a charged particle?

The strength of the magnetic field has a direct impact on the motion of a charged particle. A stronger magnetic field will result in a smaller radius of the particle's path, while a weaker field will result in a larger radius. Additionally, a stronger field will cause the particle to move at a higher velocity compared to a weaker field.

5. Can a charged particle experience a net force in a magnetic field?

No, a charged particle cannot experience a net force in a magnetic field. This is because the force it experiences is always perpendicular to its motion, causing it to move in a circular or helical path. However, the direction of the force can change, causing the particle to change its direction of motion.

Similar threads

  • Introductory Physics Homework Help
Replies
31
Views
947
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
726
  • Introductory Physics Homework Help
Replies
4
Views
744
  • Introductory Physics Homework Help
Replies
6
Views
253
  • Introductory Physics Homework Help
Replies
2
Views
961
  • Introductory Physics Homework Help
Replies
16
Views
320
  • Introductory Physics Homework Help
Replies
2
Views
572
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
975
Back
Top