An interesting EMI problem

Here is a problem :
Here is the solution : First we will calculate the magnetic field at any random point inside the cylinder due to rotation. Because rotating charges creates current
We will take a elemental ring of width dxdx by moving xx dA=2πRdxdA=2πRdx dq=σ2πRdxdq=\sigma 2πRdx I=Qω2πI=\frac{Q \omega }{2π}
I=μ0σRω\boxed{I=\mu_{0} \sigma R \omega } dB=+μ0σR3ωdx2(x2+R2)1.5\int dB=\int_{- \infty}^{+\infty} \frac{\mu_{0} \sigma R^{3} \omega dx}{2(x^{2}+R^{2})^{1.5}}
After integration B=μ0σRωB= \mu_{0} \sigma R \omega
This is basically a induced magnetic field due to rotating of charges if cylinder .
now due to this rotation, there will an electric field induced E=r2dBdtE= \frac{r}{2} \frac{dB}{dt} E=a2μ0σRωdtE=\frac{a}{2} \frac{\mu_{0} \sigma R \omega }{dt } Now due to acting of electric field in ring, there will act a torque in ring τ=qEa\tau=qEa τ=qaaμ0σRω2dt\tau=qa \frac{a \mu_{0} \sigma R \omega }{2dt} τdt=qaaμ0σRω2\tau dt=qa \frac{a \mu_{0} \sigma R \omega }{2}
Let , due to this torque,there will be an instant maximum angular velocity ω0\omega_{0} in the ring τdt=Iringω0\tau dt =I_{ring} \omega_{0} Iringω0=qaaμ0σRω2I_{ring} \omega _{0} =qa \frac{a \mu_{0} \sigma R \omega }{2}
Now from here maximum angular velocity ω0\omega _{0} comes as ω0=μ0qσRω2m\boxed{\omega_{0}=\frac{\mu_{0}q \sigma R \omega }{2m}}

Feel free to ask anything regarding the solution and problem.

If you have another trick/method to solve this particular problem please share it in below coments.
Thanks in advance.

Note by Talulah Riley
6 months, 4 weeks ago

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@Steven Chase @Karan Chatrath @Uros Stojkovic @Hosam Hajjir @Krishna Karthik , in case you guys are interested.

Talulah Riley - 6 months, 4 weeks ago

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