A conductor with a length of L = 15 cm is suspended horizontally on two weightless

A conductor with a length of L = 15 cm is suspended horizontally on two weightless threads in a magnetic field with an induction of B = 60 mT, and the lines of induction are directed upwards perpendicular to the conductor. A current was sent through the conductor. Current I = 2A. With what force (Fa) the magnetic field acts on the conductor?

L = 15 cm = 0.15 m.

I = 2 A.

B = 60 mT = 0.06 T.

∠α = 90 °.

Fump -?

A conductor with a current in a magnetic field is acted upon by the Ampere force Famp, the value of which is determined by the formula: Famp = I * B * L * sinα, where I is the current in the conductor, B is the magnetic induction, L is the length of the conductor, ∠α is the angle between the direction of the current and the vector of magnetic induction B.

Famp = 2 A * 0.06 T * 0.15 m * sin90 ° = 0.018 N.

Answer: a magnetic field acts on a conductor with a current with a force of Famp = 0.018 N.