A thin disc of radius R with uniform surface charge density sigma is rotating with angular velocity omega about its central axis. What is the magnetic moment of the disc?

A. 14 R^4 — Consider a ring element of radius r and width dr. The charge on this ring is dq = sigma * 2 * pi * r * dr, and its current is dI = dq * omega / (2 * pi). The magnetic moment is the integral of dI * pi * r^2.

JEE Main · PhysicsHard

$A thin disc of radius R with uniform surface charge density σ is rotating with angular velocity ω about its central axis. What is the magnetic moment of the disc?$

A. $\frac{1}{4} π σ ω R^{4}$
B. $\frac{1}{2} π σ ω R^{4}$
C. $\frac{2}{3} π σ ω R^{4}$
D. $\frac{1}{8} π σ ω R^{4}$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{1}{4} π σ ω R^{4}$

Solution

d μ = (σ 2 π r d r) \frac{ω}{2 π} (π r^{2}) = π σ ω r^{3} d r . Integrating from 0 to R gives \frac{1}{4} π σ ω R^{4} .

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