The period of revolution of a particle in a circle in a uniform magnetic field, whose induction 10 ^ 3 T is 3.14 * 10 ^ -12 s.

| November 20, 2020 | education

The period of revolution of a particle in a circle in a uniform magnetic field, whose induction 10 ^ 3 T is 3.14 * 10 ^ -12 s. Determine the speed with which the particle flew into the magnetic field, if before that it passed the potential difference of 1000 V.

Having passed the potential difference U, the particle with charge q and mass m acquired kinetic energy E:
E = mv² / 2 (1).
This energy is numerically equal to the work of the field:
A = qU (2);
mv² / 2 = qU (3);
v = root (2U * (q / m)) (4).
The ratio q / m can be found knowing the period of revolution of a particle T in a magnetic field with induction B:
T = (2 * pi * m) / qB (5);
qBT = 2 * pi * m;
q / m = 2 * pi / BT (6);
Substitute the expression for q / m from (6) into (4):
V = root ((2U * 2 * pi) / TB) = root ((2 * 1000 V * 2 * 3.14) / (1000 T * 3.14 * 10⁻¹² s)) = 2 * 10⁶ m / from.
Answer: V = 2 * 10⁶ m / s.



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