A conductor weighing 50 g and a length of 20 cm, suspended on two threads, is in equilibrium in a magnetic field

| February 24, 2021 | education

A conductor weighing 50 g and a length of 20 cm, suspended on two threads, is in equilibrium in a magnetic field with an induction of 0.5 T. After a current was passed through it, the thread tension increased by 2 times. What is the current strength?

m = 50 g = 0.05 kg.

g = 10 N / kg.

L = 20cm = 0.2m.

B = 0.5 T.

N2 = 2 * N1.

I -?

m * g = 2 * N1 – condition of equilibrium of a conductor without current.

m * g + Fа = 2 * N2 – the condition of equilibrium of the conductor through which the current flows.

The Ampere force Fа is expressed by the formula: Fа = I * B * L.

m * g + I * B * L = 2 * 2 * N1.

I * B * L = 4 * m * g / 2 – m * g.

The current in the conductor I will be determined by the formula: I = m * g / B * L.

I = 0.05 kg * 10 N / kg / 0.5 T * 0.2 m = 5 A.

Answer: the current in the conductor was I = 5 A.



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