A hollow, right circular cone of base radius R and height h , with...

For a point on the axis at a distance $z \gg R$ and $z \gg h$, the rotating cone can be considered as a magnetic dipole. The magnetic field on the axis of a dipole is given by:

$B = \dfrac{\mu_0}{4\pi} \dfrac{2M}{z^3}$

where $M$ is the magnetic dipole moment of the cone.

To find $M$, consider an elemental ring on the cone at a vertical distance $y$ from the origin, with vertical thickness $dy$. The radius of this ring is $r = \dfrac{R}{h}y$ and its slant length is $dl = \dfrac{\sqrt{R^2+h^2}}{h}dy$.

The surface charge density of the cone is $\sigma = \dfrac{Q}{\pi R \sqrt{R^2+h^2}}$.

The charge on the elemental ring is:

$dq = \sigma (2\pi r dl) = \dfrac{Q}{\pi R \sqrt{R^2+h^2}} \left(2\pi \dfrac{R}{h}y \dfrac{\sqrt{R^2+h^2}}{h}dy\right) = \dfrac{2Q}{h^2}y dy$

The current due to the rotation of this ring is:

$dI = \dfrac{dq}{T} = \dfrac{dq \omega}{2\pi} = \dfrac{Q\omega}{\pi h^2}y dy$

The magnetic moment of this elemental ring is:

$dM = dI (\pi r^2) = \left(\dfrac{Q\omega}{\pi h^2}y dy\right) \pi \left(\dfrac{R}{h}y\right)^2 = \dfrac{Q\omega R^2}{h^4}y^3 dy$

Integrating from $y = 0$ to $y = h$, the total magnetic moment is:

$M = \int_{0}^{h} \dfrac{Q\omega R^2}{h^4}y^3 dy = \dfrac{Q\omega R^2}{h^4} \left[\dfrac{y^4}{4}\right]_{0}^{h} = \dfrac{Q\omega R^2}{4}$

Alternatively, using the gyromagnetic ratio for a uniform charge distribution, $\dfrac{M}{L_{ang}} = \dfrac{Q}{2m}$. The moment of inertia of a hollow cone is $I = \dfrac{1}{2}mR^2$, so the angular momentum is $L_{ang} = \dfrac{1}{2}mR^2\omega$. Thus, $M = \dfrac{Q}{2m} \left(\dfrac{1}{2}mR^2\omega\right) = \dfrac{1}{4}QR^2\omega$.

Substituting $M$ into the magnetic field formula:

$B = \dfrac{\mu_0}{4\pi} \dfrac{2}{z^3} \left(\dfrac{1}{4}QR^2\omega\right) = \dfrac{\mu_0}{4\pi} \dfrac{QR^2\omega}{2z^3}$

Comparing this with the given expression $B = \dfrac{n\mu_0}{4\pi} \dfrac{QR^2\omega}{z^3}$, we get:

$n = \dfrac{1}{2} = 0.5$

Answer: $0.5$