Determine the length of the active part of the conductor if a force of 0.05H acts on it in a magnetic field with

| August 13, 2021 | education

Determine the length of the active part of the conductor if a force of 0.05H acts on it in a magnetic field with an induction of 10mT. The strength of the electric current in the conductor is 50A. The field and current are mutually perpendicular.

Data: B – field induction (B = 10 mT = 10-2 T); FA – Ampere force (FA = 0.05 N); I is the current in the presented conductor (I = 50 A); sinα = 1 (field and current are perpendicular).

To find out the length of the active part of the presented conductor, consider the formula: FA = I * B * lх * sinα, from where we express: lх = FA / (I * B * sinα).

Let’s make a calculation: lх = FA / (I * B * sinα) = 0.05 / (50 * 10-2 * 1) = 0.1 m.

Answer: The length of the active part of the presented conductor is 0.1 m.



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