A wire that is 0.50 meters long and carrying a current of 8.0 A is at rights angles to a uniform magnetic...?

Basically, there is a magnetic field in a certain direction and a wire with a certain current running through it is perpendicular to the direction of this magnetic field. (Note that the current's direction is also perpendicular to the field.)
To find the strength of the magnetic field, you can use the equation:
F = ILxB
Where F is the force, I is the current, L is the length of wire carrying the current, and B is the strength of the magnetic field.
This cross product can also be written:
F = ILBsin(theta)
Where theta is the angle between the direction of the wire/current and the direction of the field. They give you this angle: It's 90 degrees. The sine of 90 degrees is just 1. So the equation becomes:
F = ILB(1) = ILB
Plug in the known values of force, length, and current, and solve for B.
0.4 N = (8 A)(0.5m)(B)
B = 0.1 N/A(m) = 0.1 Tesla

F=BiLsin(A)where a is the angle between the magnetic feild and the wire.Here A=90 0.40=0.50*8*.therefore B=0.10