In a uniform magnetic field with an induction of 0.25 T, a conductor 10 cm long and 40 kg in weight is horizontally located

In a uniform magnetic field with an induction of 0.25 T, a conductor 10 cm long and 40 kg in weight is horizontally located. The magnetic induction lines are perpendicular to the conductor. How much current must flow through a conductor to keep it in equilibrium in a magnetic field?

L = 10 cm = 0.1 m.

m = 40 kg.

∠α = 90 °.

g = 9.8 m / s2.

B = 0.25 T.

I -?

The force of gravity Ft, which acts on the conductor, is balanced by the force of Ampere Fа: Ft = Fа.

The force of gravity is determined 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 * B * L * sinα, where I is the current in the conductor, B is the magnetic induction of the field, L is the length of the conductor, ∠α is the angle between the direction of the current and the magnetic induction.

m * g = I * B * L * sinα.

I = m * g / B * L * sinα.

I = 40 kg * 9.8 m / s2 / 0.25 T * 0.1 m * sin90 ° = 15680 A.

Answer: in order for the conductor to be in equilibrium, the current strength in it must be I = 15680 A.