A straight conductor with a length of 30 cm and a mass of 1.2 g, the current strength in which 4 A

| March 25, 2021 | education

A straight conductor with a length of 30 cm and a mass of 1.2 g, the current strength in which 4 A is in equilibrium in a uniform magnetic field, the induction modulus of which is 20 mT. Determine the angle between the direction of the current and the horizontal lines of induction.

Data: mpr – the mass of the taken rectilinear conductor (mpr = 1.2 g; in the SI system mpr = 1.2 * 10 ^ -3 kg); l is the length of the conductor (l = 30 cm; in the SI system l = 0.3 m); I – current (I = 4 A); B – induction modulus (B = 20 mT or B = 0.02 T).

Const: g – acceleration due to gravity (g ≈ 10 m / s2).

To find out the angle between the direction of the current and the induction lines, consider the equality: FA (Ampere force) = Fт (gravity) and I * B * l * sinα = mпр * g, whence we express: α = arcsin (mпр * g / (I * B * l)).

Let’s calculate: α = arcsin (mpr * g / (I * B * l)) = arcsin (1.2 * 10 ^ -3 * 10 / (4 * 0.02 * 0.3)) = arcsin 0.5 = 30º.

Answer: The angle between the direction of the current and the induction lines, according to the calculation, is 30º.



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