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Question

(D) When a conducting ring rotates in a uniform magnetic field $B$ about a point $O$ on its periphery,every small element of the ring acts as a motion

Vedclass · Accepted Answer

No current flows

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(D) When a conducting ring rotates in a uniform magnetic field $B$ about a point $O$ on its periphery,every small element of the ring acts as a motional $EMF$ source. The induced $EMF$ in a rod of length $l$ rotating about one end is given by $\varepsilon = \frac{1}{2} B \omega l^2$. For any point $P$ on the ring,the velocity vector $\vec{v}$ is perpendicular to the radius vector $\vec{r}$ from the pivot $O$. The induced $EMF$ $d\varepsilon$ in an element $dl$ is $d\varepsilon = (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$. Since the magnetic field is uniform and the ring is a closed conducting loop,the net $EMF$ around the entire closed loop is $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi}{dt}$. Because the area of the ring within the magnetic field remains constant as it rotates about a point on its periphery,the magnetic flux $\Phi$ linked with the ring does not change. Therefore,$\frac{d\Phi}{dt} = 0$,which implies that the net induced $EMF$ in the closed loop is zero. Consequently,no current flows through the ring.

$A$ conducting ring of radius $a$ is rotated about a point $O$ on its periphery as shown in the figure in a plane perpendicular to a uniform magnetic field $B$ which exists everywhere. The rotational velocity is $\omega$. Choose the correct statement$(s)$ related to the induced current in the ring.

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