A conductor with a current of 4A and a length of 50 cm in a magnetic field is exposed to a current of 0.16 N.

| August 12, 2021 | education

A conductor with a current of 4A and a length of 50 cm in a magnetic field is exposed to a current of 0.16 N. Find the magnetic induction if the conductor is perpendicular to the field lines.

I = 4 A.

L = 50cm = 0.5m.

Famp = 0.16 N.

∠α = 90.

IN – ?

A conductor that is in a magnetic field with induction B is acted upon by the Ampere force Famp, the value of which is determined by the formula: Famp = I * B * L * sinα, where I is the current in the conductor, L is the length of the conductor, ∠α is the angle between the direction of the current in the conductor and the vector of magnetic induction B.

Let us express the magnetic induction B by the formula: B = Famp / I * L * sinα.

B = 0.16 N / 4 A * 0.5 m * sin90 = 0.08 T.

Answer: the magnetic induction of the field in which the conductor is located is B = 0.08 T.



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