A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040 T at a right angle to the field. What is the radius of the proton's resulting orbit? (e = 1.60 × 10-19 C, mproton = 1.67 × 10-27 kg)

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (), as follows:

(1)

Where:

m: is the proton's mass =  1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

  (2)

By conservation of energy, we can find the velocity of the proton:

  (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

the answer is a)

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