Final answer:
The magnitude and direction of the magnetic field experienced by a current of 16 A which is perpendicular to the field, resulting in a force of 0.35 N/m, is 0.021875 T directed in the negative z direction.
Step-by-step explanation:
To calculate the magnitude and direction of the magnetic field through which a current of 16 A passes when it experiences a magnetic force per unit length of 0.35 N/m, we use the formula for the magnetic force on a length l of wire carrying a current I in a uniform magnetic field B, which is F = I × l × B × sin(θ). Since the wire is perpendicular to the magnetic field, θ = 90°, and thus sin(θ) = 1. Rearranging the equation to solve for B, we have B = F / (I × l). Given that the force per unit length is 0.35 N/m and the current is 16 A, we can calculate B as 0.35 N/m divided by 16 A, yielding a magnetic field strength of 0.021875 T. According to the right-hand rule (RHR-1) for determining the direction of the magnetic force on a current-carrying conductor, the force being in the negative y direction with the current along the positive x-axis, indicates that the magnetic field direction must be in the negative z direction.