An electric current with a force of 15 A passes through a horizontal conductor 10 cm long and weighing 4 g.

An electric current with a force of 15 A passes through a horizontal conductor 10 cm long and weighing 4 g. Find the value of the magnetic induction into which this conductor must be placed in order for its gravity to be balanced by the ampere force

L = 10 cm = 0.1 m.

m = 4 g = 0.004 kg.

g = 9.8 m / s2.

I = 15 A.

Ft = Fа.

B -?

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 Fа is expressed by the formula: Fа = I * L * B, where I is the current in the conductor, L is the length of the conductor, B is the magnetic induction of the field.

m * g = I * L * B.

The magnitude of the magnetic induction of the field B will be determined by the ratio: B = m * g / I * L.

B = 0.004 kg * 9.8 m / s2 / 15 A * 0.1 m = 0.026 T.

Answer: the conductor is located in a magnetic field with induction B = 0.026 T.