Magnetism+Viscosity+Rotation

The fluid shear drag torque should be equal to the electric torque. This whole problem consists of analyzing the contributions from infinitesimal concentric rings.

Begin fluid portion

Infinitesimal fluid shear force and torque:

$dF = \mu \, dA \, \frac{v}{d} = \mu \, 2 \pi \, r \, dr \, \frac{r \, \omega_0}{d} = \frac{2 \pi \, \omega_0 \, \mu}{d} r^2 \, dr \\ d \tau = dF \, r = \frac{2 \pi \, \omega_0 \, \mu}{d} r^3 \, dr$

Total fluid shear torque

$\tau = \int_0^R \frac{2 \pi \, \omega_0 \, \mu}{d} r^3 \, dr = \frac{2 \pi \, \omega_0 \, \mu \, R^4}{4d} = \frac{ \pi \, \omega_0 \, \mu \, R^4}{2d}$

End fluid portion

Begin electric portion

Flux linkage for loop of radius r:

$\lambda = \pi \, r^2 \, B = \pi \, r^2 \, b t$

Voltage for loop of radius r:

$\epsilon = \dot{\lambda} = \pi \, r^2 \, b$

Electric field at radius r:

$E =\frac{\epsilon}{2 \pi \, r} = \frac{1}{2} r \, b$

Infinitesimal charge at radius r:

$dQ = Q \frac{2 \pi \, r \, dr}{\pi \, R^2} = \frac{2 Q}{R^2} \, r \, dr$

Infinitesimal force at radius r:

$dF = E \, dQ = \frac{Q b}{R^2} \, r^2 \, dr$

Infinitesimal torque at radius r:

$d \tau = r \, dF = \frac{Q b}{R^2} \, r^3 \, dr$

Total electric torque:

$\tau = \int_0^R \frac{Q b}{R^2} \, r^3 \, dr = \frac{Q b \, R^2}{4}$

End electric portion

Setting torques equal

$\frac{ \pi \, \omega_0 \, \mu \, R^4}{2d} = \frac{Q b \, R^2}{4} \\ \mu = \frac{2 \, d \, Q b \, R^2}{4 \pi \, \omega_0 \, R^4} = \frac{d \, Q b }{2 \pi \, \omega_0 \, R^2} = \frac{2 \times \pi \times 2 }{2 \pi \times 10 \times 4} = \boxed{0.05}$