Explanation:

When a charged particle is accelerated through a potential difference, it gains kinetic energy. This kinetic energy can be calculated using the equation:

K.E. = qV

where K.E. is the kinetic energy, q is the charge of the particle, and V is the potential difference.

When this charged particle enters a uniform magnetic field, it experiences a magnetic force that acts perpendicular to its velocity. This force is given by the equation:

F = qvB

where F is the magnetic force, v is the velocity of the particle, and B is the magnetic field strength.

Since the magnetic force acts perpendicular to the velocity of the particle, it causes the particle to move in a circular path. The radius of this circular path can be determined using the equation:

F = mv^2/r

where m is the mass of the particle and r is the radius of the circular path.

By equating the magnetic force with the centripetal force, we can derive the following equation:

qvB = mv^2/r

Simplifying this equation, we get:

r = mv/qB

Effect of doubling the potential difference:

When the potential difference V is doubled, it means the charged particle gains twice the amount of kinetic energy. This increase in kinetic energy does not affect the magnetic force experienced by the particle, as it is dependent on the velocity and the magnetic field strength.

Therefore, the equation for the radius of the circular path remains the same:

r = mv/qB

Since the mass, charge, and magnetic field strength remain constant, doubling the potential difference does not have any effect on the radius of the circle. The charged particle will continue to move in a circle of the same radius.

Conclusion:

Doubling the potential difference does not affect the radius of the circle in which the charged particle moves. The radius of the circle is determined by the mass, charge, velocity, and magnetic field strength, but not by the potential difference.