A metal rod of length l = 20 cm, attached to two identical vertical springs of stiffness k = 10 N / m each

A metal rod of length l = 20 cm, attached to two identical vertical springs of stiffness k = 10 N / m each, is in a horizontal uniform magnetic field with induction B = 0.2 T. A current I = 5A is passed through the rod in such a direction that the deformation of the spring disappears. Find the elongation Δl of the springs after changing the direction of the current to the opposite.

L = 20cm = 0.2m.

k = 10 N / m.

B = 0.2 T.

I = 5 A.

Δl -?

Since in two cases the force of gravity acts on the conductor, which is directed vertically downward, we will not take into account its value in this problem.

In the first case, the equilibrium condition of the rod will take the form: m * g = I * B * L + 2 * k * x1, where x1 is the spring elongation in the first case.

In the second case, the equilibrium condition for the rod will take the form: m * g + I * B * L = 2 * k * x2, where x2 is the spring elongation in the second case.

Subtract the first from the first equation: m * g – m * g – I * B * L = I * B * L + 2 * k * x1 – 2 * k * x2.

2 * I * B * L = 2 * k * x2 – 2 * k * x1.

Δl = x2 – x1 = x2 – x1 = I * B * L / k.

Δl = 5 A * 0.2 T * 0.2 m / 10 N / m = 0.02 m.

Answer: after changing the direction of the current in the conductor, the change in the length of the spring was Δl = 0.02 m.