In a uniform magnetic field with an induction of 5 * 10 ^ -3 at a speed of 20 m / s, a straight wire

In a uniform magnetic field with an induction of 5 * 10 ^ -3 at a speed of 20 m / s, a straight wire 40 cm long and with a resistance of 20 ohms moves perpendicular to the field. What current would flow through the conductor if it were shorted?

Voltage in a moving conductor:
U = Ф / ∆t, where Ф – magnetic flux, ∆t – time.
The conductor moves at a constant speed: Ф = B * l * ∆s, where B is the magnetic induction (B = 5 * 10 ^ -3 T), l is the length of the wire (l = 40 cm = 0.4 m), ∆s – moving.
U = Ф / ∆t = B * l * ∆s / ∆t, where ∆s / ∆t = V (speed, V = 20 m / s).
From Ohm’s Law:
U = I * R, where I is the current strength, R is the resistance (R = 20 ohms).
Let’s calculate the current strength:
I * R = B * l * V.
I = B * l * V * / R = (5 * 10 ^ -3) * 0.4 * 20/20 = 0.002 A.
Answer: The electric current is 0.002 A.