Find Constant Value for Magnetic Field of Rotating Charged Cone

To find the magnitude of the magnetic field at a point $(0,0,z)$ along the axis of the rotating cone where $z \gg R$ and $z \gg h$, we can treat the rotating cone as a magnetic dipole with a total magnetic dipole moment $M$.

Step 1: Charge distribution on the cone

Let the apex of the hollow, right circular cone be at the origin $(0,0,0)$ and its axis lie along the $Z$-axis. At a distance $z'$ along the axis from the origin ($0 \le z' \le h$), the radius of the circular cross-section is: $r(z') = \frac{R}{h} z'$

The slant height of the cone is $L = \sqrt{R^2 + h^2}$. The total surface area of the curved surface is: $A = \pi R L$

Since the total charge $Q$ is uniformly distributed over the surface, the surface charge density $\sigma$ is: $\sigma = \frac{Q}{\pi R L}$

Consider a thin circular ring element on the cone's surface at position $z'$ of width along the slant height $dl = \frac{L}{h} dz'$. The charge $dq$ on this ring element is given by: $dq = \sigma \cdot (2\pi r) dl = \left(\frac{Q}{\pi R L}\right) \cdot 2\pi \left(\frac{R}{h} z'\right) \cdot \left(\frac{L}{h} dz'\right) = \frac{2Q}{h^2} z' dz'$

Step 2: Magnetic Moment of the Cone

As the cone rotates about the $Z$-axis with angular velocity $\omega$, the ring element forms an effective current $dI$: $dI = \frac{\omega}{2\pi} dq = \frac{\omega}{2\pi} \left(\frac{2Q}{h^2} z' dz'\right) = \frac{Q \omega}{\pi h^2} z' dz'$

The magnetic dipole moment $dM$ due to this differential ring element of area $\pi r^2$ is: $dM = dI \cdot (\pi r^2) = \left(\frac{Q \omega}{\pi h^2} z' dz'\right) \cdot \pi \left(\frac{R}{h} z'\right)^2 = \frac{Q \omega R^2}{h^4} z'^3 dz'$

Integrating $dM$ over the entire height of the cone from $z' = 0$ to $z' = h$: $M = \int_0^h \frac{Q \omega R^2}{h^4} z'^3 dz' = \frac{Q \omega R^2}{h^4} \left[ \frac{z'^4}{4} \right]_0^h = \frac{1}{4} Q \omega R^2$

Step 3: Magnetic Field along the Axis

At a far point $(0,0,z)$ on the axis where $z \gg R$ and $z \gg h$, the magnetic field is given by the axial magnetic field formula of a dipole: $B = \frac{\mu_0}{4\pi} \frac{2M}{z^3}$

Substituting the expression for $M$: $B = \frac{\mu_0}{4\pi} \frac{2 \left(\frac{1}{4} Q \omega R^2\right)}{z^3} = \frac{0.5 \, \mu_0}{4\pi} \frac{Q R^2 \omega}{z^3}$

Comparing this with the given expression: $B = \frac{n \mu_0}{4\pi} \frac{Q R^2 \omega}{z^3}$

We find the value of $n$ to be: $n = 0.5$