A particle of charge e and mass m moves with a velocity v in a magnetic field B applied perpendicular to the motion of the particle. The radius r of its path in the field is _______

CONCEPT:

• When moving through a magnetic field, the charged particle experiences a force.
• When the direction of the velocity of the charged particle is perpendicular to the magnetic field:
  • Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
  • And the particle starts to follow a curved path.
  • The particle continuously follows this curved path until it forms a complete circle.
  • This magnetic force works as the centripetal force.
• Centripetal force (F_C) = Magnetic force (F_B)

​⇒ qvB = mv²/R

​⇒ R = mv/qB

where q is the charge on the particle, v is the velocity of it, m is the mass of the particle, B is the magnetic field in space where it circles, and R is the radius of the circle in which it moves.

EXPLANATION:

Given that particle has charge e; mass = m; and moves with a velocity v in a magnetic field B. So

• Centripetal force (FC) = Magnetic force (FB)

​​⇒ qvB = mv2/R

\(\Rightarrow R=\frac{mv}{qB}\)

\(\Rightarrow r=\frac{mv}{Be}\)

So the correct answer is option 1.