Question

A soleniod 150 cm long with1550 turns has a central magnetic field of 150 mT. Find he current through the coil

Answer 1

$B=\mu_0 I\frac NL\implies I=\frac{BL}{\mu_0 N},$

$B=\frac{150\cdot 10^{-3}\cdot 1.5}{4\cdot 3.14\cdot 10^{-7}\cdot 1550}=115.5~A.$