An electron flies into a magnetic field perpendicular to the lines of magnetic induction at a speed

An electron flies into a magnetic field perpendicular to the lines of magnetic induction at a speed of 10 ^ 7 m / s. calculate the radius of curvature of the trajectory along which the electron will move if the magnetic field induction is 5.6 mT

Let’s translate all the values ​​from given to the SI system:
B = 5.6 mT = 5.6 * 10 ^ -3 T.
The Lorentz force acts on an electron in a magnetic field:
F = qe * v * B * sin α, where qe is the electron charge 1.6 * 10 ^ -19 C, v is the electron speed, B induction, α = angle to the lines of magnetic induction.
The Lorentz force according to the second law is F = me * a, a = v² / R, we have a circular motion:
F = me * v² / R, where me is the mass of the electron 9.11 * 10 ^ -31 kg, v is the speed of the electron, R is the radius of the trajectory.
Let’s equate:
qe * v * B * sin α = me * v² / R
Let us express the radius of the trajectory from this expression:
R = me * v² / (qe * v * B * sin α) = me * v / (qe * B * sin α)
Substitute the numbers:
R = me * v / (qe * B * sin α) = 9.11 * 10 ^ -31 * 10 ^ 7 / (1.6 * 10 ^ -19 * 5.6 * 10 ^ -3 * 1) = 0.01 m.
Answer: the radius of curvature of the electron trajectory is 0.01 m.