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.



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