A copper conductor with a current of 15 A, located horizontally, is suspended in air under the influence of a magnetic field.

A copper conductor with a current of 15 A, located horizontally, is suspended in air under the influence of a magnetic field. The cross-sectional area of the conductor is 1.5 mm2, the length is 1 m. The magnetic field is directed horizontally. Find the value of the magnetic induction B.

I = 15 A.

S = 1.5 mm2 = 1.5 * 10-6 m2.

L = 1 m.

g = 10 N / kg.

ρ = 8900 kg / m3.

B -?

The gravity of the copper conductor m * g is balanced by the Ampere force Famp: m * g = Famp.

Let us express these forces by the corresponding formulas: m * g = ρ * V * g = ρ * L * S * g, Famp = I * B * L.

ρ * L * S * g = I * B * L.

B = ρ * L * S * g / I * L = ρ * S * g / I.

B = 8900 kg / m3 * 1.5 * 10-6 m2 * 10 N / kg / 15 A = 8.9 * 10-3 T.

Answer: the magnitude of the magnetic field that keeps the conductor in the air is B = 8.9 * 10-3 T.