In a uniform horizontal magnetic field with an induction of 25 mT, a straight metal rod 40 cm long

In a uniform horizontal magnetic field with an induction of 25 mT, a straight metal rod 40 cm long and 60 g in weight was suspended on long weightless wires perpendicular to the magnetic lines. How much current must be passed through the rod so that the tension of the wires becomes equal to zero?

B = 25 mT = 25 * 10-3 T.

L = 40cm = 0.4m.

m = 60 g = 0.06 kg.

g = 9.8 m / s2.

∠α = 90 °.

I -?

The condition of equilibrium of the rod, according to 1 Newton’s law, is the balancing of the force of gravity Ft and the force of Ampere Famp: Ft = Famp.

The force of gravity Ft is determined by the formula: Ft = m * g.

Ampere force is expressed by the formula: Famp = I * B * L * sinα.

Since the rod is located perpendicular to the lines of force of the magnetic field, then ∠α = 90 °, sin90 ° = 1.

m * g = I * B * L.

I = m * g / B * L.

I = 0.06 kg * 9.8 m / s2 / 25 * 10-3 T * 0.4 m = 58.8 A.

Answer: current strength I = 58.8 A must pass through the rod.