Question 8.21: Because the cylinders in Ex. 8.4 are left rotating (at angul...

Because the cylinders in Ex. 8.4 are left rotating (at angular velocities \omega_{a} \text { and } \omega_{b},say),there is actually a residual magnetic field, and hence angular momentum in the fields, even after the current in the solenoid has been extinguished.If the cylinders are heavy,this correction will be negligible,but it is interesting to do the problem without making that assumption .^{22}

(a) Calculate (in terms of \omega_{a} \text { and } \omega_{b}) the final angular momentum in the fields

\text { [Define } \omega =\omega \hat{ z } \text {, so } \omega_{a} \text { and } \omega_{b} \text { could be positive or negative.] }

(b) As the cylinders begin to rotate, their changing magnetic field induces an extra azimuthal electric field, which, in turn, will make an additional contribution to the torques. Find the resulting extra angular momentum, and compare it with your result in (a) .\text { [Answer: } \left.-\mu_{0} Q^{2} \omega_{b}\left(b^{2}-a^{2}\right) / 4 \pi l \hat{ z }\right]

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(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).

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