A thin copper rod AB of length l is rotating with constant angular velocity about an axis passing through the end A and perpendicular to the plane of page in a uniform magnetic field of induction $\vec{B}$ as shown in the figure. P and Q are two points on the rod such that $A P = \frac{l}{4}$ and $A Q = \frac{l}{2} .$ If ‘V’ is the induced potential at a point on the rod, then $(V_{Q} - V_{P}) / (V_{B} - V_{A}) i s$

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$\frac{1}{8}$

$\frac{5}{24}$

$\frac{3}{16}$

$\frac{3}{4}$

answer is D.

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Detailed Solution

(4)

$\begin{array}{l} V_{P} - V_{A} = \frac{1}{2} B ω {(\frac{l}{4})}^{2}, V_{Q} - V_{A} = \frac{1}{2} B ω {(\frac{l}{2})}^{2} \\ a n d V_{B} - V_{A} = \frac{1}{2} B ω l^{2} \\ ∴ \frac{V_{Q} - V_{P}}{V_{B} - V_{A}} = \frac{{(\frac{l}{2})}^{2} - {(\frac{l}{4})}^{2}}{{(l)}^{2}} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} . \end{array}$