Circular motion of charges: A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (m_proton = 1.67 × 10^-27 kg, e = 1.60 × 10^-19C)

Question

Quizplus · Accepted Answer

The proton will experience a force due to the magnetic field given by F = qvB, where q is the charge of the proton, v is its velocity, and B is the magnetic field. This force will act as a centripetal force and keep the proton moving in a circular orbit. The centripetal force is given by F = mv^2/r, where m is the mass of the proton and r is the radius of the orbit. Equating these two expressions for F, we have:

qvB = mv^2/r

Solving for r, we get:

r = mv/qB

To find the velocity of the proton, we use the conservation of energy. Since the proton starts from rest and moves through a potential difference of 1.0 kV, its change in potential energy is qV = (1.60 × 10^-19 C)(1.0 × 10^3 V) = 1.60 × 10^-16 J. This energy is converted to kinetic energy, so we have:

(1/2)mv^2 = qV

Solving for v, we get:

v = √(2qV/m)

Substituting this expression for v into the expression we found for r, we get:

r = √(2mV/qB)

Plugging in the given values for m, V, q, and B, we get:

r = √[(2)(1.67 × 10^-27 kg)(1000 V)/(1.60 × 10^-19 C)(0.040 T)] ≈ 0.11 m

Therefore, the answer is B) 0.11 m.

Circular Motion of Charges: a Proton Starting from Rest Travels