Question 12

NUMERICALHARD

A hollow, right circular cone of base radius RR and height hh, with its tip at the origin is rotating about the Z-axis with an angular velocity ω\omega, as shown in the figure. The cone carries a total charge QQ uniformly distributed on its curved surface. The magnitude of magnetic field at a point (0,0,z)(0,0,z), where z≫Rz \gg R and z≫hz \gg h, is nμ0QR2ω4πz3\frac{n \mu_0 Q R^2 \omega}{4 \pi z^3}. The value of nn is:

Question
Correct Answer: 0.5

Detailed Solution

  1. Identify the system: We have a rotating hollow cone with surface charge QQ. This system acts as a magnetic dipole for far-off points (z≫R,hz \gg R, h).

  2. Calculate the Magnetic Dipole Moment (MM): Consider an elemental ring on the curved surface at a distance ss along the slant height from the tip.

Let L=R2+h2L = \sqrt{R^2 + h^2} be the total slant height.

The radius of the ring is r=RLsr = \frac{R}{L}s.

The charge on the ring element dq=σ(2πrds)=QπRL(2πRLs)ds=2QL2sdsdq = \sigma (2\pi r ds) = \frac{Q}{\pi R L} (2\pi \frac{R}{L}s) ds = \frac{2Q}{L^2} s ds.

The magnetic moment of this rotating ring is dM=I⋅Area=(dqω2π)(πr2)=12ωr2dqdM = I \cdot Area = (\frac{dq \omega}{2\pi}) (\pi r^2) = \frac{1}{2} \omega r^2 dq.

Substituting rr and dqdq:

dM=12ω(R2L2s2)(2QL2s)ds=QωR2L4s3dsdM = \frac{1}{2} \omega (\frac{R^2}{L^2} s^2) (\frac{2Q}{L^2} s) ds = \frac{Q \omega R^2}{L^4} s^3 ds

Integrating from s=0s = 0 to s=Ls = L:

M=∫0LQωR2L4s3ds=QωR2L4[s44]0L=14QωR2M = \int_0^L \frac{Q \omega R^2}{L^4} s^3 ds = \frac{Q \omega R^2}{L^4} \left[ \frac{s^4}{4} \right]_0^L = \frac{1}{4} Q \omega R^2

  1. Calculate Magnetic Field on Axis: For a dipole at a distance z≫z \gg dimensions of the source, the magnetic field on the axis is given by: B=μ04π2Mz3B = \frac{\mu_0}{4\pi} \frac{2M}{z^3}

Substituting the value of MM:

B=μ04π2(14QωR2)z3=0.5μ0QR2ω4πz3B = \frac{\mu_0}{4\pi} \frac{2(\frac{1}{4} Q \omega R^2)}{z^3} = \frac{0.5 \mu_0 Q R^2 \omega}{4\pi z^3}

  1. Determine nn: Comparing with the given expression B=nμ0QR2ω4πz3B = \frac{n \mu_0 Q R^2 \omega}{4 \pi z^3}, we find n=0.5n = 0.5.
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