Question

Four identical charges are fixed on the periphery of a non conducting ring of radius $R$ as shown. Ring is rotated with constant angular velocity $\omega$ wrt an axis perpendicular to pane of the ring and passing through the centre of the ring Force acting on any of the charge particle only due to the magnetic interaction between the charges is $F=\frac{{\mu }_0{q}^2{\omega }^2}{4\alpha \pi }[\sqrt{\alpha }+\frac{1}{\alpha }]$
Here $\alpha$ is an integer. Find the value of $\alpha$

Answer 1

${\overrightarrow{B}}_{AD}=\frac{{\mu }_0}{4\pi }\frac{q\left(R\omega \right)\left(1/\sqrt{2}\right)}{\left(2R^2\right)}\hat{k}$

${\overrightarrow{B}}_{AB}=\frac{{\mu }_0}{4\pi }\frac{q\left(R\omega \right)\left(1/\sqrt{2}\right)}{2R^2}\hat{k}$

${\overrightarrow{B}}_{AC}=\frac{{\mu }_0}{4\pi }\frac{q\left(R\omega \right)\left(1\right)}{4R^2}\hat{k}$

${\overrightarrow{B}}_{net}=\frac{{\mu }_0}{4\pi }\frac{qR\omega }{2R^2}[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{2}]\hat{k}$

Magnetic force is $F=qV{B}_{net}$
$\overrightarrow{F}=q\left(R\omega \right)\frac{{\mu }_0}{4\pi }\frac{\left(qRw\right)}{2R^2}(\sqrt{2}+\frac{1}{2})$

$\overrightarrow{F}=\frac{{\mu }_0}{4\left(2\right)\pi }{q}^2{\omega }^2(\sqrt{2}+\frac{1}{2})$