# A straight wire with a length of l = 25 cm and a mass of m = 50 g is placed horizontally.

A straight wire with a length of l = 25 cm and a mass of m = 50 g is placed horizontally. Its ends are connected to a current source by means of light flexible conductors. The wire is in a uniform magnetic field, the induction vector of which is oriented at an angle Ф = 30 degrees relative to the direction of the current in the wire and is equal to B = 0/98 T. How much current must be passed through the wire so that the magnetic force is in balance with the force of gravity?

l = 25 cm = 0.25 m.

m = 50 g = 0.05 kg.

B = 0.98 T.

g = 9.8 m / s2.

∠φ = 30 °.

I -?

Since by the condition of the problem Ft = Famp.

The force of gravity Ft, which acts on the conductor, is expressed by the formula: Ft = m * g.

On a conductor of length l, through which a current with a current strength I flows, in a magnetic field with induction B, the Ampere force Famp acts, the value of which is determined by the formula: Famp = I * B * l * sinφ, where ∠φ is the angle between the direction of the current in the conductor and vector of magnetic induction B.

m * g = I * B * l * sinφ.

I = m * g / B * l * sinφ.

I = 0.05 kg * 9.8 m / s2 / 0.98 T * 0.25 m * sin30 ° = 4 A.

Answer: for the equilibrium of the conductor, it is necessary to pass a current with a force of I = 4 A.

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