# A 20 cm long rectilinear conductor is placed in a uniform magnetic field with an induction of 10 mT.

**A 20 cm long rectilinear conductor is placed in a uniform magnetic field with an induction of 10 mT. The current in the conductor is 2.3 A, and a force of 4 mN acts on the conductor. what is the angle between the conductor and the magnetic induction vector?**

An Ampere force acts on a conductor in a magnetic field, which is determined by the formula:

FA = B * I * L * sin (α),

where FA is the Ampere force, B is the magnetic field induction, I is the current in the conductor, L is the length of the conductor, α is the angle between the conductor and the vector of the magnetic field induction.

From this formula we define sinα:

sinα = FA / B * I * L.

We take into account that:

B = 10 (mT) = 10 * 10-3 (T) = 10-2 (T),

L = 20 (cm) = 0.2 (m),

FA = 4 (mN) = 4 * 10-3 (N).

We get:

sin (α) = 4 * 10-3 / 10-2 * 2.3 * 0.2 = 0.4 / 0.46 = 0.8696.

Find α:

α = arcsin (0.8696) = 60.41o

Answer: α = 60.41o