# The current in a horizontally located conductor 20 cm long and weighing 4 g is 10 amperes

**The current in a horizontally located conductor 20 cm long and weighing 4 g is 10 amperes. 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 acting on the conductor from the side of the magnetic field.**

L = 20cm = 0.2m.

m = 4 g = 0.004 kg.

g = 9.8 m / s2.

I = 10 A.

∠α = 90 °.

IN – ?

On a conductor of length L, through which an electric current with a current strength I flows, in a magnetic field with induction B, the Ampere force Famp acts: Famp = I * L * B * sinα, where ∠α is the angle between the direction of the current in the conductor and the magnetic induction B …

Famp = m * g – the condition of the equilibrium of the conductor.

I * L * B * sinα = m * g.

Since the conductor is located horizontally, and the lines of magnetic induction are directed vertically, then ∠α = 90 °, sin90 ° = 1.

B = m * g / I * L.

B = 0.004 kg * 9.8 m / s2 / 10 A * 0.2 m = 0.0196 T.

Answer: the magnetic induction of the field should be B = 0.0196 T.