The conductor is in a magnetic field with an induction of 1T. Its active length is 0.1m and it is pushed out of this field with a force of 1.4N. The angle between the directions of the current and the vector of the magnetic induction of the field is 45 (degrees). In this case, the current in the conductor is?
L = 0.1 m.
B = 1 T.
Fa = 1.4 N.
∠α = 45 °.
A current-carrying conductor, in a magnetic field, is acted upon by the Ampere force Fа, the value of which is determined by the formula: Fа = I * B * L * sinα, where I is the current in the conductor, B is the magnetic induction, L is the length of the conductor, ∠α – the angle between the direction of the current and the vector of magnetic induction B.
Let us express the current strength I by the formula: I = Fа / B * L * sinα.
I = 1.4 N / 1 T * 0.1 m * sin45 ° = 20 A.
Answer: the current in the conductor is I = 20 A.
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