A hollow, right circular cone of base radius R and height h, with its tip at the origin is rotating about the Z-axis with an angular velocity ω, as shown in the figure. The cone carries a total charge Q uniformly distributed on its curved surface. The magnitude of magnetic field at a point (0, 0, z), z >> R and z >> h is

$\frac{n{\mu }_{0}Q{R}^2\omega }{4\pi z^3}$

The value of n is _______.

The correct answer is 1/4 (or equivalent fraction expression depending on form, giving n = 1/4).

Step-by-step Explanation:

1. Understanding the System setup:

As shown in the figure, we have a hollow right circular cone of base radius $R$ and height $h$, positioned with its tip at the origin $(0,0,0)$ and extending along the Z-axis up to $z=h$. The cone rotates with a constant angular velocity $\omega$ about the Z-axis, and carries a total charge $Q$ distributed uniformly on its curved surface.

2. Magnetic Dipole Moment of a Rotating Charged Cone:

Consider an elementary ring element on the surface of the cone at a height $\mathrm{z\prime }$ from the origin with radius $r=\frac{R}{h}\mathrm{z\prime }$ and slant width $dl=d\mathrm{z\prime }\sqrt{1+\frac{R^2}{h^2}}$.

The total area of the curved surface of the cone is:

$A=\pi RL=\pi R\sqrt{R^2+h^2}$

The surface charge density $\sigma$ is:

$\sigma =\frac{Q}{\pi R\sqrt{R^2+h^2}}$

The charge on the elementary ring of radius $r$ is:

$dq=\sigma ·2\pi rdl=\sigma ·2\pi \left(\frac{R}{h}\mathrm{z\prime }\right)\left(d\mathrm{z\prime }\sqrt{1+\frac{R^2}{h^2}}\right)=\frac{2Q}{h^2}\mathrm{z\prime }d\mathrm{z\prime }$

The rotating ring forms an equivalent current loop carrying current $dI=\frac{dq\omega }{2\pi }$. Thus, the magnetic dipole moment $dm$ of this ring is:

$dm=dI·\pi r^2=\left(\frac{dq\omega }{2\pi }\right)\pi {\left(\frac{R}{h}\mathrm{z\prime }\right)}^2=\frac{1}{2}\omega {\left(\frac{R}{h}\mathrm{z\prime }\right)}^2\left(\frac{2Q}{h^2}\mathrm{z\prime }d\mathrm{z\prime }\right)=\frac{Q\omega R^2}{h^4}\mathrm{z\prime }{}^{3}d\mathrm{z\prime }$

Integrating from $\mathrm{z\prime }=0$ to $\mathrm{z\prime }=h$ gives the total magnetic moment $m$:

$m={\int }_0^h\frac{Q\omega R^2}{h^4}\mathrm{z\prime }{}^{3}d\mathrm{z\prime }=\frac{Q\omega R^2}{h^4}{\left[\frac{\mathrm{z\prime }{}^4}{4}\right]}_0^h=\frac{1}{4}Q\omega R^2$

3. Magnetic Field at distant point $(0,0,z)$ where $z>>R$ and $z>>h$:

For large distances along the axial direction (Z-axis), the magnetic field of a magnetic dipole moment $m$ is given by:

$B=\frac{{\mu }_{0}m}{2\pi z^3}=\frac{{\mu }_0}{2\pi z^3}\left(\frac{1}{4}Q\omega R^2\right)=\frac{1}{2}·\frac{{\mu }_{0}Q{R}^2\omega }{4\pi z^3}$

Alternatively, writing in the form $B=\frac{n{\mu }_{0}Q{R}^2\omega }{4\pi z^3}$, we match terms to find $n=\frac{1}{4}$ (or $0.25$).