the magnetic force on a straight 0.15m segment of wire carrying a current of 4.5 A is 1.0 N. What is the magnitude of the component of the magnetic field that is perpendicular to the wire?

Answer 1

F = Bx*I*L
where Bx is the magnetic field normal to the wire.

Bx = F/(I*L) = 1.0/(4.5*0.15)
= 1.48 Tesla