A current carrying wire placed perpendicular to a magnetic field experiences a force equal to
[tex]F=ILB[/tex]
where
I is the current in the wire
L is the wire length
B is the intensity of the magnetic field
In our problem, the length of the wire is L=1.0 m, the magnetic field strength is B=0.20 T and the force exerted on the wire is F=0.60 N. If we re-arrange the equation and we plug these numbers into it, we find the current in the wire:
[tex]I= \frac{F}{LB}= \frac{0.60 N}{(1.0 m)(0.20 T)}=3 A [/tex]
