A proton that has received speed as a result of the passage of a potential difference of 1 kilovolt falls

A proton that has received speed as a result of the passage of a potential difference of 1 kilovolt falls into a magnetic field with an induction of 0.2 T, determine the radius.

ΔU = 1 kV = 1000 V.

B = 0.2 T.

q = 1.6 * 10-19 Cl.

m = 1.67 * 10-27 kg.

R -?

Let us express the speed of the proton V before it enters the magnetic field, when it has passed the accelerating potential difference according to the formula: q * ΔU = m * V2 / 2.

V2 = q * ΔU / m.

V = √ (q * ΔU / m).

V = √ (1.6 * 10-19 Cl * 1000 V / 1.67 * 10-27 kg) = 31 * 104 m / s.

A proton in a magnetic field is acted upon by the Lorentz force Fl: Fl = q * V * B.

m * a = q * V * B – 2 Newton’s law.

We express the centripetal acceleration by the formula: a = V2 / R.

m * V2 / R = q * V * B.

R = m * V2 / q * V * B = m * V / q * B.

R = 1.67 * 10-27 kg * 31 * 104 m / s / 1.6 * 10-19 C * 0.2 T = 0.0162 m.

Answer: the proton will move in a circle with a radius of R = 0.0162 m