The galvanometer has a scale limit of 200μA and an internal resistance of 80 Ohm. With the help of a shunt

The galvanometer has a scale limit of 200μA and an internal resistance of 80 Ohm. With the help of a shunt, it is supposed to measure currents up to 1 mA with this galvanometer. They want to make the shunt from a nichrome wire with a diameter of 0.4 mm. What length of wire do you need to take for this? The specific resistance of nichrome p = 1.2 Ohm × mm ^ 2 / m.

Voltage drop U on the galvanometer at current Ig = 200 μA and resistance Rg = 80 Ohm:
U = Ig * Rg.
The sum of the currents through the shunt Ish and the galvanometer Ig is equal to 1 mA:
Ig + Ish = 0.001 A;
Ish = 0.001 A – Ig.
The shunt is connected in parallel with the galvanometer, so the voltage drop across the shunt is also equal to U:
U = Ish * Rsh = (0.001 A – I₁) * Rsh;
Ig * Rg = (0.001 A – Ig) * Rsh.
Rsh = (Ig * Rg) / (0.001 A – Ig) = (0.0002 A * 80 Ohm) / (0.001 – 0.0002 A) = (0.0002 * 80 Ohm) / 0.0008 = 20 Ohm. The resistance of a nichrome wire shunt is directly proportional to the length L, specific resistance r and inversely proportional to the cross-sectional area S = (pi * d²) / 4 (d – diameter):
Rш = rL / S = rL / (pi * d² / 4) = 4rL / (pi * d²);
L = (pi * d² * Rsh) / 4r = (3.14 * 0.4² mm² * 20 Ohm) / (4 * 1.2 Ohm * mm² / m) = 2.09 m.
Answer: 2.09 m