This is how I hula hoop

First, we calculate the electric field inside the loop: $2\pi rE=\frac{d\Phi}{dt}=\pi r^2 \frac{dB}{dt}$ , so $E=\frac{dB}{dt} \frac{r}{2}$ .

The moment of electric force is: $M=Eqr=\frac{dB}{dt} \frac{qr^2}{2}$ .

We have: $dL=Mdt=dB \frac{qr^2}{2}$ or $mr^2 d\omega=dB \frac{qr^2}{2}$ .

Therefore, $\omega=\frac{qB_0}{2m}=0.005 rad/s$ .

We know that the magnetic flux passing perpendicularly to any loop is given by $\phi = BA$

Since the magnetic field is suddenly swicthed off there is change in magnetic flux passing through the loop and hence, an emf will be induced.

$e = \frac{d\phi}{dt}$ and also $e = 2\pi rE$ Plugging the value of flux into above equation and equating we get

$2\pi rE = \frac{dBA}{dt}$ ,

$E = \frac{rdB}{2dt}$

Hence , force on loop is given by $F = qE = \frac{qr^2dB}{2dt}$

Net moment on the loop is

$M = rF = I\frac{d\omega}{dt}$

You get finally $\frac{qr^2dB}{2dt} = \frac{mr^2d\omega}{dt}$

This implies $d\omega = \frac{q}{2m}dB$

Integrating both the sides you will get

$\omega = 0.005 rad/s$

$\omega=\frac{L}{I}=\frac{1}{I}\int dt M=\\ =\frac{1}{I}\int dt\int_{loop}rdF=\frac{r}{I}\int dt\int_{loop}dq E=\\ =\frac{r}{I}\frac{dq}{dl} \int dt\int_{loop}dl E=\frac{r}{I}\frac{dq}{dl} \int dt\int_{area}dS \frac{d}{dt}B=\\ =\frac{r}{I}\frac{dq}{dl} \int_{area}dS \int dt\frac{d}{dt}B=\frac{r}{I}\frac{dq}{dl} \int_{area}dS\Delta B=\\ =\frac{r}{I}\frac{dq}{dl}B_0\int_{area}dS=\frac{r}{mr^2}\frac{q}{2\pi r}B_0\pi r^2=\\ =\frac{1}{2}\frac{q}{m}B_0=\frac{1}{2}\times 10^{-1}~C/kg\times 0.1~T=0.005~rad/s$

Very easy. d(flux)/dt= E(2πr) πr². d(B)/dt= E(2πr) Therefore Torque= Eqr=say, j jdt=qr(rdB/2) Integrating both sides L=Iw= mr²w=qr²(∆B/2) w=q∆B/2m w=5÷1000=0.005