The 30 cm conductor is placed horizontally. What value should the induction of the magnetic field have so that the gravity force of a conductor with a mass of 6 g is balanced by the Ampere force? A current of 5 A flows through the conductor.
L = 30cm = 0.3m.
m = 6 g = 0.006 kg.
g = 10 m / s2.
I = 5 A.
Ft = Famp – 1 Newton’s law.
The force of gravity Ft, which acts on the conductor, is expressed by the formula: Ft = m * g, where m is the mass of the conductor, g is the acceleration of gravity.
The Ampere force Famp is expressed by the formula: Famp = I * L * B * sinα, where ∠α is the angle between the magnetic induction and the direction of the current in the conductor, I is the current in the conductor, L is the length of the conductor, V is the magnetic induction of the field.
We will assume that the conductor is parallel to the vector of magnetic induction B, therefore ∠α = 00, sin00 = 1.
Famp = I * L * V.
m * g = I * L * B.
We express the magnetic induction of the field B by the ratio: B = m * g / I * L.
B = 0.006 kg * 10 m / s2 / 5 A * 0.3 m = 0.04 T.
Answer: the induction of the magnetic field is B = 0.04 T.
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