What accelerating potential difference U must a particle with a mass m = 0.5 mg and a charge q = 2 μC

What accelerating potential difference U must a particle with a mass m = 0.5 mg and a charge q = 2 μC pass through so that in a uniform magnetic field with induction B = 5 mT a force Fl = 2 * 10 ^ -6 N would act on it? The magnetic field is directed perpendicular to the particle velocity. The initial velocity of the particle is v0 = 0.

m = 0.5 mg = 0.5 * 10 ^ -3 kg.

q = 2 μC = 2 * 10 ^ -6 C.

Fl = 2 * 10 ^ -6 N.

∠α = 90 “.

B = 5 mT = 5 * 10 ^ -3 T.

V0 = 0 m / s.

U -?

A moving charge in a magnetic field is acted upon by the Lorentz force Fl, the value of which is determined by the formula: Fl = q * B * V * sinα, where q is the magnitude of the charge, B is the magnetic induction of the field, V is the velocity of the charge, ∠α is the angle between the magnetic induction B and speed V.

Since V and B are mutually perpendicular, then sin90 “= 1.

The formula for determining the particle velocity V will take the form: V = Fl / q * B.

V = 2 * 10 ^ -6 N / 2 * 10 ^ -6 C * 5 * 10 ^ -3 T = 200 m / s.

The work of the electric field A is equal to the change in the kinetic energy of the particle ΔEk: A = ΔEk.

ΔEk = m * V ^ 2/2.

A = q * U.

m * V ^ 2/2 = q * U.

U = m * V ^ 2/2 * q.

U = 0.5 * 10 ^ -3 kg * (200 m / s) ^ 2/2 * 2 * 10 ^ -6 Cl = 5,000,000 V.

Answer: the particle has passed the potential difference U = 5,000,000 V.