Particle 1 and particle 2 have masses of m1 = 1.5×10-8 kg and m2 = 6.2×10-8 kg, but they carry the same charge q. The two particles accelerate from rest through the same electric potential difference V and enter the same magnetic field, which has a magnitude B. The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is r1 = 12 cm. What is the radius (in cm) of the circular path for particle 2?

m₁v₁²/2 =qU => v₁ =sqrt(2qU/m₁)

m₂v₂²/2 =qU => v₂=sqrt(2qU/m₂)
v₁/v₂=sqrt(2qU/m₁)/sqrt(2qU/m₂)=sqrt(m₂/m₁)
in magnetic field:
ma=qvBsinα
a=v²/R, α=90°, sin90°=1
m v²/R=qvB
R=mv/qB
R₁/R₂=(m₁v₁/qB)/ (m₂v₂/qB)=
=m₁v₁/m₂v₂=(m₁/m₂)•sqrt(m₂/m₁)=
= sqrt(m₁/m₂)
R₂=R₁•sqrt(m₂/m₁)=…

To find the radius of the circular path for particle 2, we can use the equation for the magnetic force on a charged particle moving in a magnetic field.

The magnetic force on a charged particle moving in a magnetic field is given by the equation:

F = qvB

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

For a particle moving in a circular path, the velocity can be expressed in terms of the radius of the circular path and the time it takes to complete one revolution.

v = (2πr)/T

where r is the radius and T is the period of the particle.

Since both particles are accelerating from rest and enter the same magnetic field, the potential difference V is the same for both particles. Therefore, the time period of the particles is the same.

Using these equations, we can equate the magnetic forces on particle 1 and particle 2:

qv1B = qv2B

Canceling the charge and magnetic field on both sides, we get:

v1 = v2

Now, substituting the expression for velocity in terms of radius and period, we have:

(2πr1)/T = (2πr2)/T

Canceling the 2π and T on both sides, we get:

r1 = r2

Therefore, the radius of the circular path for particle 2 is also 12 cm.

To solve this problem, we need to use the formula for the radius of a circular path of a charged particle moving through a magnetic field, which is given by:

r = (mv) / (qB)

where:
- r is the radius of the circular path
- m is the mass of the particle
- v is the velocity of the particle
- q is the charge of the particle
- B is the magnitude of the magnetic field

In this case, both particle 1 and particle 2 have the same charge q. Also, both particles start from rest and enter the same magnetic field. Therefore, the velocity of both particles will be the same.

To find the velocity, we can use the fact that the work done by the electric field is equal to the change in kinetic energy of the particles. The work done can be calculated using the formula:

W = qV

where:
- W is the work done
- q is the charge of the particle
- V is the electric potential difference

The change in kinetic energy of the particles can be calculated using the formula:

ΔKE = (1/2)mv^2

where:
- ΔKE is the change in kinetic energy
- m is the mass of the particle
- v is the velocity of the particle

Since both particles start from rest, their initial kinetic energy is zero. Therefore, the change in kinetic energy is equal to the final kinetic energy.

So, we can set up the equation:

qV = (1/2)mv^2

Simplifying this equation, we can solve for v:

v = sqrt[(2qV) / m]

Now that we have the value of v, we can substitute it into the formula for the radius to find the radius of the circular path for particle 2:

r2 = (m2v) / (qB)

Substituting the values given in the problem:

m1 = 1.5×10-8 kg
m2 = 6.2×10-8 kg
q is the same for both particles
r1 = 12 cm = 0.12 m (since the radius needs to be in meters)
B is the magnitude of the magnetic field
V is the electric potential difference

Using the given values, calculate the velocity v using the equation:

v = sqrt[(2qV) / m1]

Next, substitute the calculated value of v into the formula for r2:

r2 = (m2v) / (qB)

Finally, convert the radius r2 from meters to centimeters to get the answer.