A charged particle falls perpendicularly into a uniform magnetic field and traces a circular path. The relationship between radius r of the circular path and KE of the particle is given as-

The correct answer is option 4) i.e. r ∝ \(\sqrt{KE}\)

CONCEPT:

• Magnetic force: Magnetic force is the force experienced by electrically charged particles moving in a magnetic field.
  • The magnitude of the magnetic force (F) on a charge (q) moving at a speed (v) in a magnetic field of strength B is given by: 

​\(F = qvBsin θ\)

Where θ is the angle enclosing v and B. 

• Right-hand rule: The magnetic force acting on a moving charged particle is always perpendicular to the plane formed by v and B and follows the right-hand rule.

• When a charged particle moves perpendicular to the magnetic field (θ = 90∘), it follows a curved path and undergoes circular motion.
• Here, the magnetic force supplies the centripetal force which keeps the particle in a circular motion.

Magnetic force, \(F = qvB =\frac{mv^2}{r}\) 

​Where m is the mass of the particle, v is the velocity of the particle, and r is the radius of the circular path traced by the particle.

EXPLANATION:

The kinetic energy of a particle is given by \(KE = \frac{1}{2}mv^2\)

Where m is the mass and v is its velocity.

\(\Rightarrow v = \sqrt{\frac{2KE}{m}}\)      ----(1)

The magnetic force experienced by the particle, \(F = \frac{mv^2}{r} = qvB\)

\( \Rightarrow r = \frac{mv}{qB}\)      ----(2)

Substituting (1) in (2),

\( \Rightarrow r = \frac{m{\sqrt{\frac{2KE}{m}}}}{qB} = \frac{\sqrt{2mKE}}{qB}\)

\(\Rightarrow r \propto \sqrt{KE}\)