A straight conductor 20 cm long is placed in a uniform magnetic field with an induction of 1.26 mT

| January 12, 2021 | education

A straight conductor 20 cm long is placed in a uniform magnetic field with an induction of 1.26 mT perpendicular to the lines of magnetic induction. Determine the force acting on the conductor if the current strength in it is 50A

I = 50 A
B = 1.26 mT = 1.26 * 10-3 T.
Famp = 0.2 N.
∠α = 900.
L = 20cm = 0.2m.
Fump -?
The ampere force Famp acts on the conductor through which the electric current flows in a uniform magnetic field, the value of 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 field, L is the active conductor length, ∠α is the angle between the direction of the current and the magnetic induction.
Famp = 50 A * 1.26 * 10-3 T * 0.2 m * sin900 = 0.0126 N.
Answer: Ampere force Famp = 0.0126 N. acts on a conductor in a magnetic field.



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