Motion of Charged Particle in a Magnetic Field

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The motion of a charged particle in a magnetic field can be proven to be circular by applying Newton's Second Law and the Lorentz force equation. When the particle's velocity is in the i direction and the magnetic field is in the -k direction, the cross product results in a force that is always perpendicular to both the velocity and the magnetic field. This perpendicular force acts as a centripetal force, leading to circular motion. The relationship between the radius of the circular path, velocity, charge, and magnetic field strength is established as r = mv/(qB). Thus, the trajectory of the charged particle must indeed be circular under these conditions.
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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.
 
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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...
 
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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.
 
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