Question

A circular wire loop of radius R rotates with an angular speed $\omega$ in a uniform magnetic field B, as shown in the figure. If the emf $\epsilon$ induced in the loop is ${\epsilon }_0$ sin $\omega t,$ then the angular speed of the loop is

• A. ${\epsilon }_{o}R/B$
• B. $2\pi {\epsilon }_o/R$
• C. ${\epsilon }_o/\left(B\pi R^2\right)$
• D. ${\epsilon }_0^2\left(B{R}^2\right)$

Answer 1

The correct option is A ${\epsilon }_{o}R/B$

A circular wire loop of radius $=R$ rotate with an

angular speed$=\omega$

in a uniform magnetic field B,

If EMF$=ϵ$ induced in the loop

${ϵ}_0\sin \omega tϵ={ϵ}_0\sin \omega t$

$ϵ=\frac{d\varphi }{dt}=\frac{d}{dt}\left({ϵ}_0\sin \omega t\right)={ϵ}_0\cos \omega t\times \omega =-\frac{\Delta \left(Blx\right)}{\Delta t}ϵ={ϵ}_0\omega \cos \omega t\varphi =\int BdA=\int B\times \pi R^2=\int \frac{{ϵ}_0\cos \omega t\times \omega }{lx}\times \pi R^2=\int \frac{{ϵ}_0\cos \omega t\times \omega dt}{2R\times dx}=\int \frac{{ϵ}_0\cos \omega t\times \omega }{2R}=\frac{{ϵ}_{0}R}{B}\frac{x}{t}=\frac{{ϵ}_{0}R}{B}$