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A frame ABCD is rotating with an angular velocity $$\omega$$ about an axis passing through the point O perpendicular to the plane of paper as shown in fig 4.212. A uniform magnetic field $$\bar B$$ is applies into the plane of paper in the region as shown below. Match the following.
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Correct option is A. 2,3,6,7,9,11,12,15
As we know $$e=\dfrac{1}{2} B\omega l^2$$then,
i. $$e_{OA}=\dfrac {1}{2} B \omega (OA)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2$$
ii. $$e_{OD}=\dfrac {1}{2} B \omega (OD)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2$$
iii.$$e_{OC}=\dfrac {1}{2} B \omega L^2$$ or $$e_{CD}=E_{OD}-E_{OC}=\dfrac {1}{2} B \omega L^2 $$
iv. $$e_A-e_O=B\omega L^2$$
$$e_D-e_O=B\omega L^2$$
$$e_A-e_D=0$$
i. $$e_{OA}=\dfrac {1}{2} B \omega (OA)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2$$
ii. $$e_{OD}=\dfrac {1}{2} B \omega (OD)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2$$
iii.$$e_{OC}=\dfrac {1}{2} B \omega L^2$$ or $$e_{CD}=E_{OD}-E_{OC}=\dfrac {1}{2} B \omega L^2 $$
iv. $$e_A-e_O=B\omega L^2$$
$$e_D-e_O=B\omega L^2$$
$$e_A-e_D=0$$
ii. $$e_{OD}=\dfrac {1}{2} B \omega (OD)^2=\dfrac {1}{2} B \omega (\sqrt {2}L)^2=B \omega L^2$$
iii.$$e_{OC}=\dfrac {1}{2} B \omega L^2$$ or $$e_{CD}=E_{OD}-E_{OC}=\dfrac {1}{2} B \omega L^2 $$
iv. $$e_A-e_O=B\omega L^2$$
$$e_D-e_O=B\omega L^2$$
$$e_A-e_D=0$$
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