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Q.

Fig shows L-shaped rod rotating about its end O
in a plane perpendicular to the magnetic field B. The part OA of the rod is non-conducting while the part AB is conducting. The induced e.m.f. between the ends A and B is

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a

${ωBL}^{2}$

b

$\frac{1}{2} {ωB}^{2} L$

c

$2 {ωBL}^{2}$

d

$\frac{1}{2} {ωBL}^{2}$

answer is A.

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Detailed Solution

Consider a small element dx as shown in fig.(22). Here (dx x v) ll B

Question Image

$\begin{array}{l} ∴ e_{AB} = \int_{0}^{L} [dxωrsin θ] B \\ = ωB \int_{0}^{L} xdx [∵ rsin θ = x] \\ = \frac{1}{2} {ωBL}^{2} \end{array}$

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