Question

A particle moving in a magnetic field increases its velocity, then its radius of the circle
A. Decreases
B. Increases
C. Remains the same
D. Becomes half

Answer 1

Hint: When a charged particle enters into a region of magnetic field of uniform magnetic field strength then due to magnetic field interaction with moving charge there is magnetic force applied on the moving charge. The magnetic force is given by Lorentz's law of force.


Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.



Complete answer:
Let the mass of the particle is m and the charge is q.
If the magnetic field strength in the region is B then the magnetic force acting on the particle is,
\[\vec F = q\left( {\vec v \times \vec B} \right)\]
The vertical component of the magnetic force will cause the path of the motion of the charged particle as circular.
The deflection of the particle will be in the direction of the magnetic force and hence the magnetic force will act as centripetal force.
If the speed of the particle is v, then the centrifugal force will be,
\[{F_C} = \dfrac{{m{v^2}}}{r}\], here r is the radius of the circular path.
At radial equilibrium state, the outward force is equal to the inward centripetal force,
\[\dfrac{{m{v^2}}}{r} = qvB\]
\[r = \dfrac{{mv}}{{Bq}}\]
So, the radius of the circular path is proportional to the speed of the particle.
Hence, when we increase the speed of the particle then the radius of the particle will also increase.

Therefore, the correct option is (B).

Note:The radius of the circular path depends on the speed of the particle and not the velocity, so keeping the magnitude of the velocity constant changing the direction will not have any effect on the radius of the circular path.