A wire carries a steady current of 2.4 A. A straight section of the wire is 0.75 m long and lies along the x-axis within a uniform field B = (1.6k) T. If the current is in the positive x-direction, what is the magnetic force on the section of wire.
Solution:
The direction of LxB is the -j
direction. Since L and
B are perpendicular to each other
F = ILB = (2.4 A)(0.75 m)(1.6 T) = 2.88 N.
The force on the section of wire is F = -2.88 Nj.