Question

An iron bar 23cm with 0.12kg mass is placed perpendicular to a magnetic field B = 8 · 10−2T.

Determine direction and intensity of the current which has to travel through the bar such that the bar is suspended in air.

Answer 1

$F_A=IBl,$

$F=mg,$

$F_A=F,\implies$

$I=\frac{mg}{Bl},$

$I=\frac{0.12\cdot 9.8}{8\cdot10^{-2}\cdot 0.23}=64~A.$

The current is directed along the bar.