Answer:
Approximately [tex]17.8\; {\rm L}[/tex].
Explanation:
Consider a straight wire section of length [tex]L[/tex] carrying a current of [tex]I[/tex]. If that wire is placed in a magnetic field of strength [tex]B[/tex] at an angle of [tex]\theta[/tex] relative to the field, the magnitude of the magnetic force on this wire would be:
[tex]F = B\, I\, L\, \sin(\theta)[/tex].
In this question, [tex]\theta = 90^{\circ}[/tex] since the wire is perpendicular to the magnetic field. The magnitude of the magnetic force on this wire would be:
[tex]\begin{aligned}F &= B\, I\, L\, \sin(\theta) \\ &= (0.250\; {\rm T}) \, (15.0\; {\rm A}) \, (4.75\; {\rm m})\, (\sin(90^{\circ}) \\ &\approx 17.8\; {\rm N} \end{aligned}[/tex].
