In a magnetic field with an induction of 0.2 mT, a particle flies in at a speed of 10 ^ 7 m / s

| September 23, 2021 | education

In a magnetic field with an induction of 0.2 mT, a particle flies in at a speed of 10 ^ 7 m / s perpendicular to the lines of magnetic induction. Determine the radius of the circle along which the particle moves, if it is a proton, and find the frequency of its rotation.

To find the values ​​of the radius of the circle along which the specified proton moves, and the frequency of its rotation, we will use the equality: F = Fl; mp * a = mp * V ^ 2 / R = qp * V * B, whence we express: R = mp * V ^ 2 / (qp * V * B) = mp * V / (qp * B) and V = 2 * Π * R * ν, whence we express: ν = V / (2 * Π * R).

Constants and variables: mp – proton mass (mp = 1.673 * 10-27 kg); V is the speed of the proton (V = 10 ^ 7 m / s); qp – proton charge = 1.602 * 10 ^ -19 C; B – field induction (B = 0.2 mT = 0.2 * 10 ^ -3 T).

Calculation: a) The radius of the circle of the proton: R = mp * V / (qp * B) = 1.673 * 10 ^ -27 * 10 ^ 7 / (1.602 * 10 ^ -19 * 0.2 * 10 ^ -3) = 522 , 2 m.

b) Proton rotation frequency: ν = V / (2 * Π * R) = 10 ^ 7 / (2 * 3.14 * 522.2) = 3049.3 Hz.

Answer: The radius of the circle of the indicated proton is 522.2 m; rotation frequency – 3049.3 Hz.



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