An electron moves in a uniform magnetic field with an induction of 10 mT. At some point in time

An electron moves in a uniform magnetic field with an induction of 10 mT. At some point in time, the vector of its velocity, equal to 1Mm / s, makes an angle of 30 degrees with the direction of the magnetic field. Calculate the radius and pitch of the helix along which the electron moves.

B = 10 mT = 0.01 T.

q = 1.6 * 10 ^ -19 Cl.

∠α = 30 °.

V = 1 Mm / s = 1 * 10 ^ 6 m / s.

m = 9.1 * 10 ^ -31 kg.

R -?

d -?

The electron will move in a spiral with radius R and step d.

We decompose the electron velocity V into two components: Vх = V * cosα, Vу = V * sinα

The electron is acted upon by the Lorentz force Fl: Fl = q * V * B * sinα.

m * a = q * V * B * sinα – 2 Newton’s law.

We express the centripetal acceleration by the formula: a = Vу ^ 2 / R = (V * sinα) ^ 2 / R.

m * (V * sinα) 2 / R = q * V * B * sinα.

R = m * V * sinα / q * B.

R = 9.1 * 10 ^ -31 kg * 1 * 10 ^ 6 m / s * sin30 ° / 1.6 * 10 ^ -19 C * 0.01 T = 0.000284 m.

d = Vx * T = V * cosα * T, where T is the time of one complete revolution of the electron.

T = 2 * P * R / Vу = 2 * P * R / V * sinα.

T = 2 * 3.14 * 0.000284 m / 1 * 10 ^ 6 m / s * sin30 ° = 3.56 * 10 ^ -9 s.

d = 1 * 10 ^ 6 m / s * cos30 ° * 3.56 * 10 ^ -9 m / s = 3 * 10 ^ -3 m.

Answer: the electron will move in a spiral with a radius of R = 0.000284 m and a step d = 3 * 10 ^ -3 m.