What is the induction of the magnetic field in which a force of 50 mN acts on a conductor with an active part length of 5 cm?

| May 29, 2021 | education

What is the induction of the magnetic field in which a force of 50 mN acts on a conductor with an active part length of 5 cm? The conductor is located perpendicular to the magnetic induction vector.

l = 5 cm = 0.05 m. Famp = 50 mN = 0.05 N. I = 25 A. ∠α = 90 “. B -?
An Ampere force acts on a conductor with a current in a magnetic field, 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 magnetic field, l is the length of the conductor, ∠α is the angle between the direction of the current I in the conductor and the magnetic induction B. B = Famp / I * l * sinα. B = 0.05 N / 25 A * 0.05 m * sin90 “= 0.04 T.
Answer: the magnetic induction of the field is B = 0.04 T.



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