If the work required to move the conductor of length $L$ shown in figure, one full turn in the positive direction at a rotational frequency  $N$ revolutions per minute, if  $\overrightarrow{B}=B_0\widehat{a_r}$ ( $B_0$ is positive constant and  $\widehat{a_r}$ is a unit vector in radial direction), is  $-y\pi r{B}_{0}IL$ . Then  $y$ is  

The magnetic force on the conductor is  ${\overrightarrow{F}}_B=I\overrightarrow{L}\times \overrightarrow{B}=IL\widehat{k}\times B_0\widehat{a_r}=B_{0}IL\left(\widehat{a_{\varphi }}\right)$

External force is  ${\overrightarrow{F}}_{ext}=B_{0}IL(-\widehat{a_{\varphi }})$

Where $\widehat{a_{\varphi }}$  is a unit vector in tangential direction. 
Therefore work done required to turn the conductor in one full revolution is 
$W=\underset{0}{\overset{2\pi }{\int }}{B}_{0}IL(-\widehat{a_{\varphi }}).rd\varphi \widehat{a_{\varphi }}=-2\pi r{B}_{0}IL$

The negative sign shows that field does work. The fact that work is done around a closed path shows that the force is non – conservative in this case.