The 30 cm conductor is placed horizontally. What value should the induction of the magnetic field have so that the gravity

The 30 cm conductor is placed horizontally. What value should the induction of the magnetic field have so that the gravity force of a conductor with a mass of 6 g is balanced by the Ampere force? A current of 5 A flows through the conductor.

L = 30cm = 0.3m.

m = 6 g = 0.006 kg.

g = 10 m / s2.

I = 5 A.

B -?

Ft = Famp – 1 Newton’s law.

The force of gravity Ft, which acts on the conductor, is expressed by the formula: Ft = m * g, where m is the mass of the conductor, g is the acceleration of gravity.

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

We will assume that the conductor is parallel to the vector of magnetic induction B, therefore ∠α = 00, sin00 = 1.

Famp = I * L * V.

m * g = I * L * B.

We express the magnetic induction of the field B by the ratio: B = m * g / I * L.

B = 0.006 kg * 10 m / s2 / 5 A * 0.3 m = 0.04 T.

Answer: the induction of the magnetic field is B = 0.04 T.