A positively charged particle of mass 5.20 10-8 kg is traveling due east with a speed of 30 m/s and enters a 0.37-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 4.20 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field.

Question????

To determine the magnitude of the charge on the particle, we can use the equation for the magnetic force on a charged particle:

F = qvb

Where:
- F represents the magnetic force
- q represents the charge on the particle
- v represents the velocity of the particle
- b represents the magnetic field strength

Since the particle is moving perpendicular to the magnetic field, the magnetic force will provide the centripetal force needed to keep the particle in a circular path. Therefore, we can equate the magnetic force to the centripetal force:

F = mv²/r

Where:
- m represents the mass of the particle
- v represents the velocity of the particle
- r represents the radius of the circular path

We know that the particle travels one-quarter of a circle. Therefore, the angle traversed is 90 degrees or π/2 radians. The radius of the circular path can be determined using the formula for the radius of a circular motion:

r = v*t

Where:
- v represents the velocity of the particle
- t represents the time it takes to traverse the one-quarter of a circle

Using the given values, we can calculate the radius of the circular path.