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(D) The induced electromotive force $(EMF)$ across a rotating rod of length $l$ in a magnetic field $B$ with angular velocity $\omega$ is given by $\v

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

Vedclass · Answer

(D) The induced electromotive force $(EMF)$ across a rotating rod of length $l$ in a magnetic field $B$ with angular velocity $\omega$ is given by $\varepsilon = \frac{1}{2} B \omega l^2$.For any point $X$ on the ring,the potential difference relative to the pivot point $O$ is $V_X - V_O = \frac{1}{2} B \omega r_{OX}^2$,where $r_{OX}$ is the straight-line distance from $O$ to $X$.For points $P$ and $R$,the distance from $O$ is $r_{OP} = r_{OR} = \sqrt{a^2 + a^2} = \sqrt{2}a$.Thus,$V_P - V_O = V_R - V_O = \frac{1}{2} B \omega (\sqrt{2}a)^2 = B \omega a^2$.Since $B, \omega, a^2 > 0$,we have $V_P = V_R > V_O$.For point $Q$,the distance from $O$ is $r_{OQ} = 2a$.Thus,$V_Q - V_O = \frac{1}{2} B \omega (2a)^2 = 2 B \omega a^2$.Now,$V_Q - V_P = (V_Q - V_O) - (V_P - V_O) = 2 B \omega a^2 - B \omega a^2 = B \omega a^2$.Since $V_P - V_O = B \omega a^2$,it follows that $V_Q - V_P = V_P - V_O$.Therefore,both statements $(B)$ and $(C)$ are correct.

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$A$ copper rod $AB$ of length $L$,pivoted at one end $A$,rotates at a constant angular velocity $\omega$,at right angles to a uniform magnetic field of induction $B$. The e.m.f. developed between the midpoint $C$ of the rod and end $B$ is

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