A conducting rod weighing 200 g is held in a uniform magnetic field on smooth horizontal rails, the distance

A conducting rod weighing 200 g is held in a uniform magnetic field on smooth horizontal rails, the distance between which is 20 cm. The magnetic lines are directed vertically, the magnetic induction module is 0.1 T, the current in the rod is 10 A. What path will this rod travel in 0.5 s after it is released?

m = 200 g = 0.2 kg.

L = 20cm = 0.2m.

B = 0.1 T.

I = 10 A.

t = 0.5 s.

∠α = 90 °.

S -?

When the conductive rod is released, it will begin to move under the action of the Ampere force Famp, the value of which is expressed by the formula: Famp = 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 vector of magnetic induction B.

2 Newton’s law will have the form: m * a = I * B * L * sinα.

Let us express the acceleration a from the formula: S = a * t ^ 2/2.

a = 2 * S / t2.

m * 2 * S / t ^ 2 = I * B * L * sinα.

S = I * B * L * sinα * t ^ 2 / m * 2.

S = 10 A * 0.1 T * 0.2 m * sin90 ° * (0.5 s) ^ 2 / 0.2 kg * 2 = 0.125 m.

Answer: the rod will travel along the rails with a path S = 0.125 m.