Under the action of the forces of a uniform magnetic field, a straight conductor with a current strength I = 3 A

Under the action of the forces of a uniform magnetic field, a straight conductor with a current strength I = 3 A and a mass of m = 20 g begins to move perpendicular to the lines of magnetic induction. What magnetic flux will cross this conductor by the time when its velocity is equal to v = 10 m / s? What will be the work of the field forces on the movement of the conductor?

∠α = 90.

I = 3 A.

m = 20 g = 0.02 kg.

V0 = 0 m / s.

V = 10 m / s.

F -?

A -?

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

ΔEk = m * V ^ 2/2 – m * V0 ^ 2/2.

ΔEk = 0.02 kg * (10 m / s) ^ 2/2 – 0.02 kg * (0 m / s) ^ 2/2 = 1 J.

A = 1 J.

Ф = B * S = B * x * L, where x is the distance the conductor moved.

The ampere force Famp = I * B * L * sinα acts on a conductor with a current in a magnetic field.

A = Famp * x = I * B * L * sinα * x = I * B * S * sinα = I * Ф * sinα.

Ф = А / I * sinα.

Ф = 1 J / 3 A * sin900 = 0.34 Wb.

Answer: A = 1 J, F = 0.34 Wb.