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Magnetic field far along the axis of a spinning charged hollow cone: the rings' moments add to QR^2 w/4
JEE Advanced 2026
Paper 1 · Question 12
Physics
Magnetic moment of a spinning charged cone and its far field on the axis
Moderate
A hollow cone with charge Q on its curved surface spins about its axis. Its charge forms current rings whose moments add to m = QR^2 w/4, independent of the height, and far up the axis the field is the dipole value (mu0/4pi) 2m/z^3, so n = 0.5.
Open the interactive simulationSub-topics covered
- A rotating charge as a current: dI = dq w/2pi
- Magnetic moment of a current loop: m = I pi r^2
- Surface charge on a cone: a ring at slant fraction u has r = uR and dq = 2Qu du
- Axial field of a magnetic dipole far away: B = (mu0/4pi) 2m/z^3
What it teaches
- Each ring of the spinning cone is a current loop with moment dm = dq w r^2/2
- The moments add: m = Q R^2 w integral_0^1 u^3 du = Q R^2 w/4, whatever the height h
- Far on the axis every current distribution looks like a dipole: B = (mu0/4pi) 2m/z^3 = (mu0/4pi) Q R^2 w/(2 z^3), n = 0.5
- The axial dipole field is twice the broadside value; the charge sits mostly near the rim, so m is not Q w <r^2>/2 with a uniform weight
Simulation ID
- ADV-2026-P1-PHY-Q12 · revision 1
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