# In a horizontal conductor with a length of 50 cm and a mass of 10 g, the current strength is 20A.

**In a horizontal conductor with a length of 50 cm and a mass of 10 g, the current strength is 20A. Find the induction of the magnetic field in which you need to place the conductor so that the force of gravity is balanced by the force of Ampere.**

L = 50cm = 0.5m.

m = 10 g = 0.01 kg.

g = 10 m / s2.

I = 20 A.

B -?

The condition for the equilibrium of a conductor with a current in a magnetic field is the equality of the gravity Ft and the Ampere force Famp that act on it: Ft = Famp.

Let us express the force of gravity Ft and the force of Ampere Famp according to the formulas: Ft = m * g, Famp = I * B * L.

m * g = I * B * L.

B = m * g / I * L.

B = 0.01 kg * 10 m / s2 / 20 A * 0.5 m = 0.02 T = 2 * 10-2 T.

Answer: for the equilibrium of the conductor in a magnetic field, the induction of the magnetic field should be B = 2 * 10-2 T.