01The question
Q.12 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), where z ≫ R and z ≫ h, is (nμ0/4π)(QR²ω/z³). The value of n is:
02The experiment
Answer—
Press START: the cone spins, its charge makes current rings, and a probe measures B far up the axis.
Ready.
Height h
Speed
03Detailed solution
Answer: n = —
- Slice the cone into rings. At a fraction u of the way along the slant (0 at the tip, 1 at the rim), the ring has radius r = uR. The curved area up to u grows as u², so a ring of width du holds charge dq = Q · 2u du.
- Each ring is a current loop. Spinning at ω, the charge passes any point ω/2π times a second, so dI = dq ω/2π. Its magnetic moment is
dm = dI · πr² = (1/2) dq ω r² = QωR² u³ du.
- Add them up. m = QωR² ∫₀¹ u³ du = QR²ω/4. The height h drops out.
- Far along the axis, the cone looks like a dipole. For z ≫ R, h, the axial field of a dipole is
B = (μ₀/4π) · 2m/z³ = (μ₀/4π) · QR²ω/(2z³) ⇒ n = 0.5.
- What the experiment checks. The probe adds up the exact on-axis field of every ring, μ₀ dI r²/2(r² + (z − zring)²)3/2, without assuming a dipole. As z grows, Bz³ settles at the value that gives n = 0.5, for any height of cone.
Understand the concept
A rotating charged body is a stack of current loops. Far away, all that survives of their fields is the total magnetic moment. For a loop of radius r carrying charge dq at angular speed ω, the moment is dqωr²/2. The moment is not equal to the charge times the average r²: the charge sits mostly near the rim (dq ∝ u), where r² is largest, so the moment is weighted by ∫2u·u² = 1/2. On the axis, the far field of a dipole is (μ₀/4π)(2m/z³), twice its broadside value.
Key takeaways
- A spinning charge dq on a ring of radius r has magnetic moment dq ω r²/2.
- Uniform charge on a cone's curved surface: dq = 2Qu du with r = uR, which gives m = QR²ω/4.
- On-axis dipole field: B = μ₀ 2m/(4πz³), so n = 0.5. The cone's height does not matter far away.
Verification: an independent derivation, a sympy route, a numerical route (the exact Biot–Savart field of thousands of rings, and a direct surface integral), and the page's own engine, run by tests/verify_p1phyq12.py. The official key (0.50) and the MathonGo solution (0.5) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. The model uses Q = R = ω = 1 and μ₀/4π = 1, and n is dimensionless.
How to use this simulation
- Press ▶ START EXPERIMENT. The cone spins, and its surface charge circulates as current.
- Watch the rings light up from tip to rim as their magnetic moments are added. The total reaches QR²ω/4.
- The probe then climbs the z-axis. The plot shows Bz³, in units of (μ₀/4π)QR²ω, settling at 0.5.
- Change Height h: near the cone the field changes, but far away n is still 0.5.
- Replay runs it again. Reset clears it. Classroom fills the screen for a projector.