# 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 $dx$ by moving $x$ $dA=2πRdx$ $dq=\sigma 2πRdx$ $I=\frac{Q \omega }{2π}$
$\boxed{I=\mu_{0} \sigma R \omega }$ $\int dB=\int_{- \infty}^{+\infty} \frac{\mu_{0} \sigma R^{3} \omega dx}{2(x^{2}+R^{2})^{1.5}}$
After integration $B= \mu_{0} \sigma R \omega$
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 $E= \frac{r}{2} \frac{dB}{dt}$ $E=\frac{a}{2} \frac{\mu_{0} \sigma R \omega }{dt }$ Now due to acting of electric field in ring, there will act a torque in ring $\tau=qEa$ $\tau=qa \frac{a \mu_{0} \sigma R \omega }{2dt}$ $\tau dt=qa \frac{a \mu_{0} \sigma R \omega }{2}$
Let , due to this torque,there will be an instant maximum angular velocity $\omega_{0}$ in the ring $\tau dt =I_{ring} \omega_{0}$ $I_{ring} \omega _{0} =qa \frac{a \mu_{0} \sigma R \omega }{2}$
Now from here maximum angular velocity $\omega _{0}$ comes as $\boxed{\omega_{0}=\frac{\mu_{0}q \sigma R \omega }{2m}}$

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
5 months, 1 week ago

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## Comments

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

- 5 months, 1 week ago

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