JEE Advanced 2026 — Paper 1 — PHY-III solution

JEE Advanced2026Paper 1PHY-III

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 $\frac{n μ _{0}}{4 π} \frac{Q R ^{2} ω}{z ^{3}}$ . The value of $n$ is:

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Solution

Treat the rotating charged cone as a magnetic dipole. For a charged body in pure rotation, the magnetic moment relates to the angular momentum by $\frac{M}{L} = \frac{q}{2 m}$ for a uniform charge-to-mass ratio, giving $M = \frac{Q}{2 m} \cdot I ω$ , where $I$ is the moment of inertia of an equivalent mass distribution on the cone's surface.

Moment of inertia of a thin conical shell about its symmetry axis: $I = \frac{1}{2} m R^{2}$ . So $M = \frac{Q}{2 m} \cdot \frac{1}{2} m R^{2} ω = \frac{Q R ^{2} ω}{4}$ .

Field on the dipole axis far away. $B_{axial} = \frac{μ _{0}}{4 π} \cdot \frac{2 M}{z ^{3}} = \frac{μ _{0}}{4 π} \cdot \frac{2}{z ^{3}} \cdot \frac{Q R ^{2} ω}{4} = \frac{μ _{0}}{8 π} \cdot \frac{Q R ^{2} ω}{z ^{3}}$ .

Comparing with $\frac{n μ _{0}}{4 π} \cdot \frac{Q R ^{2} ω}{z ^{3}}$ : $\frac{n}{4 π} = \frac{1}{8 π} \Rightarrow n = \frac{1}{2} = 0.50$ .

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