Question

A proton moves at 7.50×10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength?

a) 0.05 T
b) 0.10 T
c) 0.20 T
d) 0.40 T

Answer 1

The strength of the magnetic field that causes a proton to travel in a circular path with a radius of 0.800 m and a velocity of 7.50×107 m/s is approximately 0.10 T.

Step-by-step explanation:

The question involves calculating the strength of a magnetic field that causes a proton to move in a circular path with a given radius and velocity. The radius of the circular path is 0.800 m, and the proton's velocity is 7.50×107 m/s. When a charged particle moves in a magnetic field perpendicular to the field, it experiences a centripetal force due to the magnetic field, which is given by the equation F = qvB, where F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength. The centripetal force is also given by F = mv2/r, where m is the mass of the particle, and r is the radius of the circular path.

Equating the expressions for the magnetic force and the centripetal force, we have qvB = mv2/r. Solving for B, we get B = mv/(qr). Using the known values of a proton's mass (m = 1.67×10-27 kg) and charge (q = 1.60×10-19 C), we can calculate the magnetic field strength B.

Substituting the given values into the formula, we have B = (1.67×10-27 kg × 7.50×107 m/s) / (1.60×10-19 C × 0.800 m), which calculates to approximately 0.10 T.