A horizontal conductor 25 cm long, the electrical resistance of which is 2.4 ohms, is suspended on two thin

A horizontal conductor 25 cm long, the electrical resistance of which is 2.4 ohms, is suspended on two thin vertical insulating threads in a horizontal uniform magnetic field with an induction of 0.02 T perpendicular to the lines of magnetic induction. What voltage was applied to the conductor if the total tension of the threads increased by 20 mN after the key was closed?

L = 25cm = 0.25m.

R = 2.4 ohms.

B = 0.02 T.

ΔF = 20 mN = 0.02 N.

∠α = 90 °.

U -?

According to Ohm’s law for a section of a circuit, the voltage U is determined by the product of the current strength I in the conductor by its resistance R: U = I * R.

On a conductor of length L, through which a current flows with a current strength I, in a magnetic field with induction B, the Ampere force Famp acts, the value of which is determined by the formula: Famp = I * B * L * sinα, where ∠α is the angle between the direction of the current in the conductor and vector of magnetic induction B.

The tension of the threads increased ΔF due to the Ampere force Famp: Famp = ΔF.

ΔF = I * B * L * sinα.

I = ΔF / B * L * sinα.

U = ΔF * R / B * L * sinα.

U = 0.02 N * 2.4 Ohm / 0.02 T * 0.25 m * sin90 ° = 9.6 V.

Answer: at the ends of the conductor, the voltage is U = 9.6 V.