A section of a conductor 10 cm long is in a magnetic field with an induction of 50 mT, the strength of the electric

A section of a conductor 10 cm long is in a magnetic field with an induction of 50 mT, the strength of the electric current passing through the conductor is 10 A. What movement does the conductor make in the direction of the force of 0.004J? The conductor is located perpendicular to the lines of magnetic induction

l = 10 cm = 0.1 m. B = 50 mT = 50 * 10 ^ -3 T. I = 10 A. A = 0.004 J. ∠α = 90 “.
find and S -?
Mechanical work A is determined by the formula: A = F * S, where F is the force acting on the conductor, S is the movement of the conductor under the action of the force. The ampere force acts on the conductor, which is determined by the formula: F = l * B * I * sinα. Where l is the length of the active part of the conductor, B is the magnetic induction of the field, I is the current in the conductor, ∠α is the angle between the direction of the current and the magnetic induction. A = l * B * I * sinα * S. S = A / l * B * I * sinα. S = 0.004 J / 0.1 m * 50 * 10 ^ -3 T * 10 A sin90 “= 0.08 m.
Answer: the conductor has moved to a distance of S = 0.08 m.