Explanation:
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts perpendicular to both the velocity of the particle and the magnetic field. This force causes the particle to move in a circular path.

Given:
Initial velocity (v) of the charged particle
Initial magnetic field strength (B)
Final velocity of the charged particle = 2v
Final magnetic field strength = B/2

Formula:
The force experienced by a charged particle moving in a magnetic field is given by the formula:

F = qvB sinθ

Where:
F = Magnetic force
q = Charge of the particle
v = Velocity of the particle
B = Magnetic field strength
θ = Angle between the velocity and magnetic field

Analysis:
We need to find the change in radius of the circular path when the velocity of the charged particle is doubled and the strength of the magnetic field is halved.

1. Initial radius (r1): When the particle is moving with velocity v in the initial magnetic field B, it follows a circular path. The magnetic force provides the centripetal force required for the circular motion. Therefore, we can equate the magnetic force with the centripetal force:

F = qvB = mv²/r1

Where:
m = Mass of the particle

2. Final radius (r2): When the velocity of the charged particle is doubled (2v) and the strength of the magnetic field is halved (B/2), the magnetic force acting on the particle changes. We can equate the new magnetic force with the centripetal force:

F = q(2v)(B/2) = m(2v)²/r2

Solution:
By comparing the two equations, we can find the relationship between the initial and final radii:

qvB = m(2v)²/r1
q(2v)(B/2) = m(2v)²/r2

Simplifying the equations:

r1 = 2mv/qB
r2 = 4mv/qB

Taking the ratio of r2 and r1:

r2/r1 = (4mv/qB)/(2mv/qB)
r2/r1 = 4/2
r2/r1 = 2

Therefore, the radius becomes 2 times larger when the velocity of the charged particle is doubled and the strength of the magnetic field is halved. Hence, the correct answer is option 'c' - 2 times.