Question

The magnetic field at the centre of a uniform charged ring (Q) of radius $R$ rotating at angular velocity $\omega$ about an axis passing through a diameter is $\frac{{\mu }_0}{4\pi }\frac{nQ\omega }{R}$, then n is

Answer 1

$dI=\frac{2dq}{T}=\frac{\omega }{\pi }\lambda Rd\theta$
$dB=\frac{{\mu }_0}{4\pi }\frac{2dI\pi R^{2}co{s}^2\theta }{R^3}$
$B=\int dB=\frac{{\mu }_0}{4\pi }\frac{2\pi }{R}\frac{\omega }{\pi }\lambda R{\int }_{-\pi /2}^{\pi /2}$
$B=\int dB=\frac{{\mu }_0}{4\pi }\frac{2\pi }{R}\frac{\omega }{\pi }\lambda R{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}co{s}^2\theta d\theta$
$B=\frac{{\mu }_0}{4\pi }\frac{Q\omega }{2R}$