A proton and an electron with masses mпр = 1.67 * 10-27 kg mel = 9.1 * 10-31 kg, accelerated by the same potential
A proton and an electron with masses mпр = 1.67 * 10-27 kg mel = 9.1 * 10-31 kg, accelerated by the same potential difference, fly into a uniform magnetic field in the direction perpendicular to the induction lines. Find the ratio of the radii of curvature of the trajectories of the proton and the electron.
mp = 1.67 * 10 ^ -27 kg.
me = 9.1 * 10 ^ -31 kg.
qp = 1.6 * 10 ^ -19 Cl.
qe = 1.6 * 10 ^ -19 Cl.
Ve = Vp.
∠α = 90 °.
Re / Rп -?
The electric charge q, which moves at a speed V in a magnetic field with induction B, is acted upon by the Lorentz force Fl, the value of which is determined by the formula: Fl = q * V * B * sinα, where ∠α is the angle between the direction of motion of the charge V and the magnetic induction vector IN.
Fle = qe * V * B * sinα.
Fлп = qп * V * B * sinα
The electron and proton will move in a spiral with radius R and step d.
mp * ap = qp * V * B * sinα, me * ae = qe * V * B * sinα – 2 Newton’s law for the electron and proton.
We express the centripetal acceleration by the formula: ap = V ^ 2 / Rp, ae = V ^ 2 / Re.
mp * V ^ 2 / Rp = qp * V * B * sinα, me * V ^ 2 / Re = qe * V * B * sinα.
Rp = mp * V ^ 2 / qp * V * B * sinα = mp * V / qp * B * sinα.
Re = me * V ^ 2 / qe * V * B * sinα = me * V / qe * B * sinα.
Rp / Re = mp * V * qe * B * sinα / me * V * qp * B * sinα = mp / me.
Rp / Re = 1.67 * 10 ^ -27 kg / 9.1 * 10 ^ -31 kg = 1835.
Answer: the radius of the circle of the proton will be 1835 times the radius of the circle of the electron.