A positively charged particle of mass 5.60 10-8 kg is traveling due east with a speed of 60 m/s and enters a 0.49-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 1.20 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.
To find the magnitude of the charge on the particle, we can use the formula for the magnetic force acting on a charged particle moving perpendicular to a magnetic field. The formula is given by:
F = qvB
where F is the magnetic force, q is the charge on the particle, v is the velocity of the particle, and B is the magnetic field strength.
In this case, the particle is moving perpendicular to the magnetic field, so the force acting on it is caused by the magnetic field and must supply the centripetal force necessary to keep it moving in a circle.
The centripetal force is given by the equation:
F = mv^2/r
where m is the mass of the particle, v is the velocity, and r is the radius of the circular path.
Since the particle travels one-quarter of a circle, the radius of the circle can be found using the formula for the circumference of a circle:
C = 2πr
where C is the circumference of the one-quarter of a circle. Therefore, r = C/(2π), where C = 1/4 of the circumference.
Finally, we can equate the magnetic force and the centripetal force:
qvB = mv^2/r
The mass of the particle and the velocity are given in the problem. We know the radius can be calculated using the circumference equation. The only unknown is the charge (q), which we need to solve for.
Rearranging the equation to solve for q, we get:
q = (mv)/rB
Plug in the values for the mass (5.60 * 10^-8 kg), velocity (60 m/s), radius (calculated using the given information), and magnetic field strength (0.49 T) to find q.
This will give you the magnitude of the charge on the particle.