Magnetism

Concept:A spinning charged cone behaves like a magnetic dipole. The magnetic field far along its axis is given by $B = \frac{\mu_0}{4\pi} \frac{2M}{z^3}$, where $M$ is the net magnetic dipole moment of the cone.Explanation:Let the cone have tip at origin, base radius $R$, height $h$, slant length $L = \sqrt{R^2 + h^2}$, and semi‑vertical angle $\alpha$ with $\sin\alpha = R/L$. Total charge $Q$ is uniformly spread over the curved surface area $\pi R L$. Take a thin ring at slant distance $l$ from tip, width $dl$. Its radius $r = l\sin\alpha$, vertical coordinate $y = l\cos\alpha$. Area of ring $dA = 2\pi r\, dl = 2\pi (l\sin\alpha)\, dl$. Surface charge density $\sigma = Q/(\pi R L)$. Charge on ring: $dq = \sigma\, dA = \frac{Q}{\pi R L} \cdot 2\pi l\sin\alpha\, dl = \frac{2Q}{L^2} l\, dl$. When cone rotates with angular speed $\omega$, the ring’s charge moves in a circle of radius $r$, giving an effective current $dI = \frac{dq}{T} = \frac{\omega}{2\pi} dq$. Magnetic dipole moment of this ring: $dM = dI \cdot (\pi r^2) = \frac{\omega}{2\pi} dq \cdot \pi (l\sin\alpha)^2 = \frac{1}{2} \omega \sin^2\alpha\, l^2 dq$. Substitute $dq$: $dM = \frac{1}{2} \omega \sin^2\alpha\, l^2 \cdot \frac{2Q}{L^2} l\, dl = \frac{Q\omega \sin^2\alpha}{L^2} l^3 dl$. Integrate over the entire slant length: $M = \int_0^L \frac{Q\omega \sin^2\alpha}{L^2} l^3 dl = \frac{Q\omega \sin^2\alpha}{L^2} \cdot \frac{L^4}{4} = \frac{1}{4} Q\omega \sin^2\alpha\, L^2$. Since $\sin^2\alpha\, L^2 = R^2$, we get $M = \frac{1}{4} Q\omega R^2$. At a far point $(0,0,z)$ with $z \gg R,h$, the axial magnetic field is $B = \frac{\mu_0}{4\pi} \frac{2M}{z^3} = \frac{\mu_0}{4\pi} \frac{2\left(\frac{1}{4} Q\omega R^2\right)}{z^3} = \frac{1}{2} \frac{\mu_0}{4\pi} \frac{Q R^2 \omega}{z^3}$. This matches the given form $n \frac{\mu_0}{4\pi} \frac{Q R^2 \omega}{z^3}$ with $n = 0.5$. Answer:$n = 0.5$