A section of a conductor 20 cm long is in a magnetic field with an induction of 50 mT. The strength of the electric

A section of a conductor 20 cm long is in a magnetic field with an induction of 50 mT. The strength of the electric current passing through the conductor is 5 A. What movement will the conductor make in the direction of the Ampere force if the work of this force is 0.005 J? The conductor is located perpendicular to the lines of magnetic induction.

To find the displacement of the taken section of the conductor, we will use the formula: A = Fa * Δr = I * B * l * sinα * Δr, whence we express: Δr = A / (I * B * l * sinα).

Variables: A – work of the Ampere force (A = 0.005 J); I is the current in the conductor (I = 5 A); B – magnetic field induction (B = 50 mT = 0.05 T); l is the length of the conductor section (l = 20 cm = 0.2 m); α is the angle between the section of the conductor and the lines of magnetic induction (α = 90º).

Calculation: Δr = A / (I * B * l * sinα) = 0.005 / (5 * 0.05 * 0.2 * sin 90º) = 0.1 m.

Answer: The section of the conductor should have moved 0.1 m.