If the work required to move the conductor shown in figure, one full turn in the positive direction at a rotational frequency `N` revolutions per minute, if `vec(B)=B_(0)hat(a)_(r)(B_(0)` is positive constant and `hat(a)_(r)` is a unit vector in radial direciton `) ,` is `y pi r B_(0)il`. Then `y` is .

Answer

2

Explanation

The magnetic force on the conductor is
`vec(F)_(B)=Ivec(L)xxvec(B)=ILhat(k)xxB_(0)hat(a)_(r)=B_(0)IL(hat(a)_(phi))`
External force is
`vec(F)_(ext)=B_(0)IL(-hat(a)_(phi))`
Where `hat(a)_(phi)` is a unit vector in tangential direction. Therefore work done requuired to turn the conductor in one full revolution is `:`
`W=int_(0)^(2pi)B_(0)IL(-hat(a)_(phi)).rdphihat(a)_(phi)=-2pirB_(0)IL`
The negative sign shows that field work. The fact that work is down around a closed path shows that the force is non`-` conservative in this case.