JEE Advanced 2026 · Paper 1, Physics question 12

Far away on the axis, the spinning cone acts as a magnetic dipole: $B=\dfrac{\mu_0}{4\pi}\dfrac{2m}{z^3}$.

Dipole moment. Let $\ell$ be the slant height; the surface charge density is $\sigma=\dfrac{Q}{\pi R\ell}$. A ring at slant distance $s$ from the tip has radius $\rho=\frac{sR}\ell$ and charge $\sigma\,2\pi\rho\,ds$, so it carries current $\sigma\rho\omega\,ds$ and has moment
$$dm=\sigma\omega\pi\rho^3\,ds.$$
Integrating:
$$m=\sigma\pi\omega\frac{R^3}{\ell^3}\int_0^\ell s^3\,ds=\frac{\sigma\pi\omega R^3\ell}4=\frac{QR^2\omega}4.$$

Field.
$$B=\frac{\mu_0}{4\pi}\frac{QR^2\omega}{2z^3},$$
so $n=0.5$.