Question

proton of charge +1.60×10⁻¹⁹ C and mass 1.67×10⁻²⁷kg is introduced into a region of B = 0.140 T with an initial velocity of 1.33×106 m/s perpendicular to B. What is the radius of the proton's path?

Answer 1

The radius of the proton's path in a magnetic field is found using the formula R = mv/(qB). With the given values of the proton's mass, charge, velocity, and the magnetic field strength, the calculated radius is approximately 1.12 meters.

Step-by-step explanation:

When a charged particle like a proton of charge +1.60×10⁻¹¹ C and mass 1.67×10⁻²⁷ kg is introduced into a region with a magnetic field perpendicular to its velocity, it will move in a circular path. To find the radius of this path, we use the formula R = mv/(qB), where m is the mass of the proton, v is its velocity, q is its charge, and B is the magnetic field strength.

Given the initial velocity v = 1.33×10⁶ m/s and the magnetic field strength B = 0.140 T, we can plug in the values to calculate the radius:

R = (1.67×10⁻²⁷ kg × 1.33×10⁶ m/s) / (1.60×10⁻¹¹ C × 0.140 T).

After calculating, we find that the radius R is approximately 1.12 meters, which is the path the proton will follow in the magnetic field.