JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Magnetic Effects of Current

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{QR ^{2} ω}{z ^{3}}

. The value of

n

is:

Answer: 0.5

Solution

The rotating charged cone can be treated as a magnetic dipole at large distances.

For a rotating charged system, the magnetic moment is related to angular momentum by:

\frac{M}{L} = \frac{q}{2 m}

The angular momentum of the rotating cone:

L = \frac{1}{2} m R^{2} ω

Therefore, magnetic moment:

M = \frac{Q}{2 m} \cdot \frac{1}{2} m R^{2} ω = \frac{Q ω R ^{2}}{4}

Magnetic field on axis of dipole at distance z:

B = \frac{μ _{0}}{4 π} \frac{2 M}{z ^{3}} = \frac{μ _{0}}{4 π} \frac{2 \cdot Q ω R ^{2} /4}{z ^{3}} = \frac{μ _{0}}{8 π} \frac{Q R ^{2} ω}{z ^{3}}

Comparing with given form:

n = \frac{1}{2} = 0.5

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Magnetic Effects of Current

Solution

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