When an alpha particle with mass m, charge 2e enters into a magnetic field B with velocity v perpendicular to the direction of the magnetic field, the radius of the curved path will be ____________.

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 (FC) = Magnetic force (FB)

​q v B = 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:

• When a charge moves through a magnetic field in a perpendicular direction, it will move in a circular motion as explained above with a radius 

​R = mv/qB

​For alpha particle according to the question mass is m, velocity is v, the charge is 2e, and the magnetic field is B. So

\( R={mv \over qB}={mv \over (2e)B}\)

• So the correct answer is option 2.