(a) The rotating shell at radius b produces a solenoidal magnetic field:
B =\mu_{0} K \hat{ z }, \text { where } K=\sigma_{b} \omega_{b} b \text {, and } \sigma_{b}=-\frac{Q}{2 \pi b l} . \text { So } B =-\frac{\mu_{0} \omega_{b} Q}{2 \pi l} \hat{ z }(a<s<b) .
(Note that if angular velocity is defined with respect to the z axis, then \omega_{b} is a negative number.) The shell at a also produces a magnetic field \left(\mu_{0} \omega_{a} Q / 2 \pi l\right) \hat{ z }, \text { in the region } s<a , so the total field inside the inner shell is
B =\frac{\mu_{0} Q}{2 \pi l}\left(\omega_{a}-\omega_{b}\right) \hat{ z },(s<a) .
Meanwhile, the electric field is
E =\frac{1}{2 \pi \epsilon_{0}} \frac{\lambda}{s} \hat{ s }=\frac{Q}{2 \pi \epsilon_{0} l s} \hat{ s }, \quad(a<s<b) .
g =\epsilon_{0}( E \times B )=\epsilon_{0}\left(\frac{Q}{2 \pi \epsilon_{0} l s}\right)\left(-\frac{\mu_{0} \omega_{b} Q}{2 \pi l}\right)(\hat{ s } \times \hat{ z })=\frac{\mu_{0} \omega_{b} Q^{2}}{4 \pi^{2} l^{2} s} \hat{ \phi } ; \quad \ell= r \times g =\frac{\mu_{0} \omega_{b} Q^{2}}{4 \pi^{2} l^{2} s}( r \times \hat{ \phi }) .
\text { Now } r \times \hat{ \phi }=(s \hat{ s }+z \hat{ z }) \times \hat{ \phi }=s \hat{ z }-z \hat{ s }, \text { and the } \hat{ s } term integrates to zero, so
L =\frac{\mu_{0} \omega_{b} Q^{2}}{4 \pi^{2} l^{2}} \hat{ z } \int d \tau=\frac{\mu_{0} \omega_{b} Q^{2}}{4 \pi^{2} l^{2}} \pi\left(b^{2}-a^{2}\right) l \hat{ z }=\frac{\mu_{0} \omega_{b} Q^{2}\left(b^{2}-a^{2}\right)}{4 \pi l} \hat{ z } .
(b) The extra electric field induced by the changing magnetic field due to the rotating shells is given by
E 2 \pi s=-\frac{d \Phi}{d t} \Rightarrow E =-\frac{1}{2 \pi s} \frac{d \Phi}{d t} \hat{\phi} , and in the region a < s < b
\Phi=\frac{\mu_{0} Q}{2 \pi l}\left(\omega_{a}-\omega_{b}\right) \pi a^{2}-\frac{\mu_{0} Q \omega_{b}}{2 \pi l} \pi\left(s^{2}-a^{2}\right)=\frac{\mu_{0} Q}{2 l}\left(\omega_{a} a^{2}-\omega_{b} s^{2}\right) ; E (s)=-\frac{1}{2 \pi s} \frac{\mu_{0} Q}{2 l}\left(a^{2} \frac{d \omega_{a}}{d t}-s^{2} \frac{d \omega_{b}}{d t}\right) \hat{\phi} .
In particular,
E (a)=-\frac{\mu_{0} Q a}{4 \pi l}\left(\frac{d \omega_{a}}{d t}-\frac{d \omega_{b}}{d t}\right) \hat{\phi}, \quad \text { and } E (b)=-\frac{\mu_{0} Q}{4 \pi l b}\left(a^{2} \frac{d \omega_{a}}{d t}-b^{2} \frac{d \omega_{b}}{d t}\right) \hat{\phi} .
The torque on a shell is N = r \times q E =q s E \hat{ z } , so
N _{a}=Q a\left(-\frac{\mu_{0} Q a}{4 \pi l}\right)\left(\frac{d \omega_{a}}{d t}-\frac{d \omega_{b}}{d t}\right) \hat{ z } ; \quad L _{a}=\int_{0}^{\infty} N _{a} d t=-\frac{\mu_{0} Q^{2} a^{2}}{4 \pi l}\left(\omega_{a}-\omega_{b}\right) \hat{ z } .
N _{b}=-Q b\left(-\frac{\mu_{0} Q}{4 \pi l b}\right)\left(a^{2} \frac{d \omega_{a}}{d t}-b^{2} \frac{d \omega_{b}}{d t}\right) \hat{ z } ; \quad L _{b}=\int_{0}^{\infty} N _{b} d t=\frac{\mu_{0} Q^{2}}{4 \pi l}\left(a^{2} \omega_{a}-b^{2} \omega_{b}\right) \hat{ z } .
L _{ tot }= L _{a}+ L _{b}=\frac{\mu_{0} Q^{2}}{4 \pi l}\left(a^{2} \omega_{a}-b^{2} \omega_{b}-a^{2} \omega_{a}+a^{2} \omega_{b}\right) \hat{ z }=-\frac{\mu_{0} Q^{2} \omega_{b}}{4 \pi l}\left(b^{2}-a^{2}\right) \hat{ z } .
Thus the reduction in the final mechanical angular momentum (b) is equal to the residual angular momentum in the fields (a).