On a straight conductor 1 m long with a current of 200A, located in a uniform magnetic field at an angle of 30 degrees

On a straight conductor 1 m long with a current of 200A, located in a uniform magnetic field at an angle of 30 degrees in the direction of the magnetic field, a force of 10 Newtons acts. Find the induction of the magnetic field and its strength.

L = 1 m.

∠α = 30 °.

Famp = 10 N.

I = 200 A.

V – ?

H -?

On a conductor with a current, in a magnetic field, the Ampere force Famp acts, 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 of the field, L is the length of the conductor, ∠α is the angle between the direction of the current and the vector of magnetic induction B.

The magnetic induction B of the field is expressed by the formula: B = Famp / I * L * sinα.

B = 10 N / 200 A * 1 m * sin30 ° = 0.1 T.

The magnetic field strength H is expressed by the formula: H = B / μ0 * μ, where μ0 is the absolute magnetic permeability, μ is the magnetic permeability of the medium.

For air μ = 1.

μ0 = 4 * P * 10-7 H / m.

H = 0.1 T / 4 * 3.14 * 10-7 H / m = 79617.8 A / m.

Answer: the conductor is in a magnetic field with induction B = 0.1 T, strength H = 79617.8 A / m.