Answer
Final answer:
Force exerted by each conductor is $F= 11.055\ N$ (or in the american system units $ F= 2.485\ lb$)
Work Step by Step
- conductors (an electric cable/wire) is attached to the armature of a DC motor. Their length is 6 inches
- the current across the armature; therefore, across the conductors is $i=90 A$
- the magnetic field density surrounding the conductors is $B=5.2\times10^{-4}Wb/in^{2}$
- We are required to find the force exerted by the conductors (cable/wire)
$$%$$
The magnetic force across a straight line is
$$F=i\ (I\times B)=i l B\ sin\theta$$
we assume that the current across the conductors and the magnetic field are perpendicular; thus, $\theta=90 $ degrees. Therefore,
$$F=i l B\ sin\theta=i l B\ sin(90) = i l B\ (1) =i l B$$
First, lets convert to SI units to make calculations easier
$$l = 6\ in=6\times 2.54 \times 10^{-2}=0.1524 \ m$$
$$B=5.2\times 10^{-4} \ Wb/in^{2}=5.2\times 10^{-4}\times \frac{1}{(2.54\times 2.54^{-2})^{2}} =0.806 \ Wb/m^{2}=0.806 \ T$$
Now, plug the values into the force equation
$$F=i l B=90\times 0.1524 \times 0.806 = 11.055\ N$$
Convert back from SI units to American units
$$F= 11.055\ \times .2248 = 2.485\ lb$$
