A wire carries 20.0 A through a 1.50 T field, perpendicular to it, with 5.00 cm of its length inside the field. What force acts on it?

1. Use F = IlB sin(theta) for a straight wire in a uniform field. 2. The current is perpendicular to the field, so theta = 90 degrees and sin(theta) = 1. 3. Convert the length: 5.00 cm = 0.0500 m. 4. Substitute: F = (20.0 A)(0.0500 m)(1.50 T)(1). 5. Multiply: (20.0)(0.0500) = 1.00, and (1.00)(1.50) = 1.50 N. 6. So the force is 1.50 N, which is a significant push on such a short length of wire. 7. An answer of 15.0 N follows from leaving the length in centimetres. _Source: OpenStax College Physics (CC BY 4.0), Ch 22 "Magnetism", section 22.7 Magnetic Force on a Current-Carrying Conductor_