Answer:
Approximately [tex]0.0575\; {\rm V}[/tex].
Explanation:
Let [tex]B[/tex] denote the magnetic field strength. Let [tex]L[/tex] denote the length of this wire. Let [tex]v[/tex] denote the speed of this wire.
Let [tex]\theta[/tex] denote the angle between the magnetic field and the motion of this wire. To find the voltage ([tex]{\rm emf}[/tex]) induced in this wire, apply the formula:
[tex](\text{emf}) = B\, L\, (v\, \sin(\theta))[/tex].
In this question, it is given that [tex]B = 0.25\; {\rm T}[/tex], [tex]L = 1\; {\rm m}[/tex], and [tex]v = 0.23\; {\rm m\cdot s^{-1}}[/tex]. Additionally, [tex]\theta = 90^{\circ}[/tex] since the magnetic field is perpendicular to the motion of the wire. Therefore, the voltage induced in this wire would be:
[tex]\begin{aligned}(\text{emf}) &= B\, L\, (v\, \sin(\theta)) \\ &= (0.25\; {\rm T})\, (1\; {\rm m}) \, (0.23\; {\rm m\cdot s^{-1}}\, (\sin(90^{\circ}))) \\ &= 0.0575\; {\rm V}\end{aligned}[/tex].
