# A conductor 0.5 m long, through which a current of 2A flows, can freely move around the frame.

**A conductor 0.5 m long, through which a current of 2A flows, can freely move around the frame. Through an insulator, this conductor is attached to a spring with a stiffness of 5 N / m. When the magnetic field is turned on, the induction vector of which is perpendicular to the plane of the frame, the spring is stretched by 20 cm. What is the field induction?**

L = 0.5 m.

I = 2 A.

k = 5 N / m.

∠α = 90 °.

x = 20 cm = 0.2 m.

IN – ?

A current-carrying conductor in a magnetic field is acted upon by the Ampere force Famp, which is balanced by the spring force Fup: Famp = Fup.

Ampere force Famp is determined by the formula: Famp = I * B * L * sin∠α∠α

The elastic force Fup is determined by Hooke’s law: Fup = k * x.

I * B * L * sinα = k * x.

The magnetic induction of the field B will be determined by the formula: B = k * x / I * L * sinα.

B = 5 N / m * 0.2 m / 2 A * 0.5 m * sin90 ° = 1 T.

Answer: the magnetic field has an induction of B = 1 T.