Consider ring like element of disc of radius r and thickness dr. If σ is charge per unit area, then charge on the element dq=σ(2πrdr) current ' i ' associated with rotating charge dq is i=2π(dq)w=σwrdr Magnetic field dB at center due to element dB=2rμ0i=2μ0σωdr $\begin{aligned}
& \mathrm{B}{\text {net }}=\int \mathrm{dB}=\frac{\mu_0 \sigma \omega}{2} \int_0^{\mathrm{R}} \mathrm{dr}=\frac{\mu_0 \sigma \omega \mathrm{R}}{2} \
& \Rightarrow \quad \mathrm{B}{\text {net }}=\frac{\mu_0 \mathrm{Q} \omega}{2 \pi \mathrm{R}} \quad\left[\because \mathrm{Q}=\sigma \pi \mathrm{R}^2\right]
\end{aligned}SoifQandwareunchangedthen\mathrm{B}{\mathrm{net}} \propto \frac{1}{\mathrm{R}}HencevariationofB{\text {net }}withR$ should be a rectangular hyperbola as represented in (A). 
A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane…
JEE Main 2012 — Physics Electromagnetism
A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure
Held on 30 Apr 2012 · Verified 6 Jul 2026.
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