Physics - Magnetism Question with Solution | TestHub

Divide the disc into rings. A ring of radius x and width dx carries charge dq = σ·2πx dx and rotates with angular velocity ω.

Equivalent current of the ring: $dI=\frac{\omega }{2\pi }dq=\sigma \omega xdx$

Field of this ring at distance y on its axis: $dB=\frac{{\mu }_{0}dI{x}^2}{2(x^2+y^2{)}^{3/2}}=\frac{{\mu }_0\sigma \omega }{2}\frac{x^{3}dx}{(x^2+y^2{)}^{3/2}}$

Integrate from x = 0 to x = r:$B=\frac{{\mu }_0\sigma \omega }{2}{\int }_0^r\frac{x^{3}dx}{(x^2+y^2{)}^{3/2}}$

Using $\int \frac{x^{3}dx}{(x^2+y^2{)}^{3/2}}=\sqrt{x^2+y^2}+\frac{y^2}{\sqrt{x^2+y^2}}$

$B=\frac{{\mu }_0\sigma \omega }{2}\left(\frac{r^2+2y^2}{\sqrt{r^2+y^2}}-2y\right)$

Hence (C).