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 , 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 n _0 4 Q R^2 z^3 . The value of n is:
A. A
For a point on the axis at a distance z R and z h, the rotating cone can be considered as a magnetic dipole. The magnetic field on the axis of a dipole is given by: B = _0 4 2M z^3 where M is the magnetic dipole moment of the cone. To find M, consider an elemental ring on the cone at a vertical distance y from the origin, with vertical thickness dy. The radius of this ring is r = R h y and its slant length is dl = R^2+h^2 h dy. The surface charge density of the cone is = Q R R^2+h^2 . The charge on the elemental ring is: dq = (2 r dl) = Q R R^2+h^2 (2 R h y R^2+h^2 h dy ) = 2Q h^2 y dy The current due to the rotation of this ring is: dI = dq T = dq 2 = Q h^2 y dy The magnetic moment of this elemental ring is: dM = dI ( r^2) = ( Q h^2 y dy ) ( R h y )^2 = Q R^2 h^4 y^3 dy Integrating from y = 0 to y = h, the total magnetic moment is: M = _ 0 ^ h Q R^2 h^4 y^3 dy = Q R^2 h^4 [ y^4 4 ]_ 0 ^ h = Q R^2 4 Alternatively, using the gyromagnetic ratio for a uniform charge distribution, M L_ ang = Q 2m . The moment of inertia of a hollow cone is I = 1 2 mR^2, so the angular momentum is L_ ang = 1 2 mR^2 . Thus, M = Q 2m ( 1 2 mR^2 ) = 1 4 QR^2 . Substituting M into the magnetic field formula: B = _0 4 2 z^3 ( 1 4 QR^2 ) = _0 4 QR^2 2z^3 Comparing this with the given expression B = n _0 4 QR^2 z^3 , we get: n = 1 2 = 0.5 Answer: 0.5
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