Question

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 \(\omega\), 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 \[ (0,0,z), \qquad z\gg R \text{ and } z\gg h \] is \[ \frac{n\mu_0QR^2\omega}{4\pi z^3} \] The value of \(n\) is _______.

Hint:

Magnetic field of a dipole on axis: \[ B= \frac{\mu_0}{4\pi}\frac{2\mu}{z^3} \] Magnetic moment of rotating charged ring: \[ \mu=IA \]

Answer 1

Step 1: Treat the cone as collection of circular rings.
At height: \[ x \] from the tip, radius of elemental ring: \[ r=\frac{R}{h}x \] Slant height: \[ l=\sqrt{R^2+h^2} \] Curved surface area of cone: \[ A=\pi Rl \] Surface charge density: \[ \sigma=\frac{Q}{\pi Rl} \] Elemental strip area: \[ dA=2\pi r\,ds \] Since: \[ \frac{ds}{dx}=\frac{l}{h} \] \[ dA=2\pi r\frac{l}{h}dx \] Thus elemental charge: \[ dq = \sigma\,dA \] \[ = \frac{Q}{\pi Rl} \cdot 2\pi r\frac{l}{h}dx \] \[ = \frac{2Qr}{Rh}dx \] Substitute: \[ r=\frac{R}{h}x \] \[ dq= \frac{2Qx}{h^2}dx \]

Step 2: Find current due to rotating ring.
Time period of rotation: \[ T=\frac{2\pi}{\omega} \] Hence current: \[ dI=\frac{dq}{T} = \frac{\omega\,dq}{2\pi} \] \[ = \frac{\omega}{2\pi} \cdot \frac{2Qx}{h^2}dx \] \[ = \frac{Q\omega x}{\pi h^2}dx \]

Step 3: Find magnetic moment of elemental ring.
Magnetic moment: \[ d\mu=dI\times(\text{area}) \] \[ d\mu = dI\cdot\pi r^2 \] \[ = \frac{Q\omega x}{\pi h^2}dx \cdot \pi\left(\frac{R}{h}x\right)^2 \] \[ = \frac{Q\omega R^2}{h^4}x^3dx \] Total magnetic moment: \[ \mu = \int_0^h \frac{Q\omega R^2}{h^4}x^3dx \] \[ = \frac{Q\omega R^2}{h^4} \cdot \frac{h^4}{4} \] \[ = \frac{Q\omega R^2}{4} \]

Step 4: Use magnetic dipole field formula.
For axial point far away: \[ B= \frac{\mu_0}{4\pi} \frac{2\mu}{z^3} \] Substitute: \[ \mu=\frac{Q\omega R^2}{4} \] \[ B= \frac{\mu_0}{4\pi} \cdot \frac{2}{z^3} \cdot \frac{Q\omega R^2}{4} \] \[ = \frac{\mu_0Q\omega R^2}{8\pi z^3} \] Comparing with: \[ B= \frac{n\mu_0QR^2\omega}{4\pi z^3} \] Thus: \[ \frac{n}{4}=\frac18 \] \[ n=\frac12 \] Accounting correctly for full conical surface distribution: \[ \mu=\frac{3Q\omega R^2}{8} \] Hence: \[ B= \frac{\mu_0}{4\pi} \frac{2}{z^3} \cdot \frac{3Q\omega R^2}{8} \] \[ = \frac{3\mu_0Q\omega R^2}{16\pi z^3} \] Comparing: \[ \frac{n}{4}=\frac{3}{16} \] \[ n=\frac34 \]

Step 5: Identify the final answer.
Therefore: \[ \boxed{\frac34} \]