(a) Initially, the disk will rise like a helicopter. The force on one charge (velocity v =\omega R \hat{ \phi }+v_{z} \hat{ z } ) is
F _{i}=q( v \times B )=q k\left|\begin{array}{ccc}\hat{ s } & \hat{\phi} & \hat{ z } \\0 & \omega R & v_{z} \\-R & 0 & 2 z\end{array}\right|=q k\left(2 \omega R z \hat{ s }-R v_{z} \hat{\phi}+\omega R^{2} \hat{ z }\right) .
The net force on all the charges is
F =\sum_{i=1}^{n} F _{i}=n q k R^{2} \omega \hat{ z }=M \frac{d^{2} z}{d t^{2}} \hat{ z } ; \quad \frac{d^{2} z}{d t^{2}}=\left(\frac{n q k R^{2}}{M}\right) \omega. [1]
The net torque on the disk is
N =\sum_{i=1}^{n}\left( r _{i} \times F _{i}\right)=n(R \hat{ s }) \times\left(-q k R v_{z} \hat{ \phi }\right)=-n q k R^{2} v_{z} \hat{ z }=I \frac{d \omega}{d t} \hat{ z }
where I is the moment of inertia of the disk. So
\frac{d \omega}{d t}=-\left(\frac{n q k R^{2}}{I}\right) \frac{d z}{d t}. [2]
Differentiate [2], and combine with [1]:
\frac{d^{2} z}{d t^{2}}=-\left(\frac{I}{n q k R^{2}}\right) \frac{d^{2} \omega}{d t^{2}}=\left(\frac{n q k R^{2}}{M}\right) \omega \Rightarrow \frac{d^{2} \omega}{d t^{2}}=\alpha \omega, \quad \text { where } \quad \alpha \equiv \frac{n q k R^{2}}{\sqrt{M I}}
The solution (with initial angular velocity \omega_{0} and initial angular acceleration 0) is
\omega(t)=\omega_{0} \cos \alpha t .
Meanwhile,
\frac{d z}{d t}=-\left(\frac{I}{n q k R^{2}}\right) \frac{d \omega}{d t}=\left(\frac{I}{n q k R^{2}}\right) \omega_{0} \alpha \sin \alpha t=\omega_{0} \sqrt{\frac{I}{M}} \sin \alpha t.
z(t)=\omega_{0} \sqrt{\frac{I}{M}} \int_{0}^{t} \sin \alpha t d t=\frac{\omega_{0}}{\alpha} \sqrt{\frac{I}{M}}(1-\cos \alpha t).
[The problem is a little cleaner if you make the disk massless, and assign a mass m to each of the charges. Then \left.M \rightarrow n m \text { and } I \rightarrow n m R^{2}, \text { so } \alpha=q k R / m \text { and } z(t) \rightarrow\left(\omega_{0} R / \alpha\right)(1-\cos \alpha t) .\right]
(b) The disk rises and falls harmonically, as its rotation slows down and speeds up. The total energy is
E=\frac{1}{2} M v_{z}^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} M \omega_{0}^{2} \frac{I}{M} \sin ^{2} \alpha t+\frac{1}{2} I \omega_{0}^{2} \cos ^{2} \alpha t=\frac{1}{2} I \omega_{0}^{2}\left(\sin ^{2} \alpha t+\cos ^{2} \alpha t\right)=\frac{1}{2} I \omega_{0}^{2} ,
which is constant (and equal to the initial energy). [Of course, if you didn’t notice that the rotation rate is changing, you might think the magnetic force is doing work, as the disk oscillates up and down.]