A rectilinear conductor with a mass of 0.02 kg and a length of 50 cm is placed in a uniform magnetic field perpendicular

A rectilinear conductor with a mass of 0.02 kg and a length of 50 cm is placed in a uniform magnetic field perpendicular to the lines of magnetic induction. what should be the induction of the magnetic field so that the conductor hangs without falling if the current in the conductor is 2 A.

m = 0.02 kg.

g = 10 m / s2.

L = 50cm = 0.5m.

∠α = 90 °.

I = 2 A.

B -?

In order for the conductor with the current to be in the air and not fall, it is necessary that the action of the forces on it be compensated. The force of gravity Ft is compensated by the Ampere force Famp: Ft = Famp.

Let us express the force of gravity Ft by the formula: Ft = m * g.

The Ampere force Famp is expressed by the formula: Famp = I * B * L * sinα, where I is the current in the conductor, B is the magnetic induction of the field, L is the length of the conductor, ∠α is the angle between the direction of the current in the conductor and the vector of magnetic induction B.

m * g = I * B * L * sinα.

B = m * g / I * L * sinα.

B = 0.02 kg * 10 m / s2 / 2 A * 0.5 m * sin90 ° = 0.2 T.

Answer: for the equilibrium of the conductor, the magnetic induction of the field should be B = 0.2 T.