Question

A non conducting disc having uniform positive charge $Q$, is rotating about its axis in anti clockwise direction with uniform angular velocity $\omega$. The magnetic field at the center of the disc is.

• A. directed outwards perpendicular to plane of disc.
• B. having magnitude $\frac{{\mu }_{0}Q\omega }{4\pi R}$
• C. directed inwards perpendicular to plane of disc.
• D. having magnitude $\frac{{\mu }_{0}Q\omega }{2\pi R}$

Answer 1

Answer: (A), (D)

having magnitude $\frac{{\mu }_{0}Q\omega }{2\pi R}$

Consider a ring of radius $x$ and thickness $dx$.
Equivalent current in this ring $=\frac{\omega }{2\pi }\times \text{charge on ring}=\frac{\omega }{2\pi }\times (2\pi x\times dx)\frac{Q}{\pi R^2}$
Magnetic field due to this ring is $dB=\frac{{\mu }_0}{2x}\left(\frac{\omega }{2\pi }\frac{2xQ}{R^2}dx\right)\therefore B={\int }_0^R\frac{{\mu }_0\omega }{2\pi }\frac{Q}{R^2}dx$
$B=\frac{{\mu }_0\omega Q}{2\pi R^2}.R=\frac{{\mu }_0\omega Q}{2\pi R}$.

Using Right hand thumb rule we can identify that magnetic field is directed outwards.

Why this Question? Tip: We need to choose element in such a way that we should be able to write B due to it. Eg.- In this case we took element as a ring since we know how to write B for a ring.