$A$ metallic rod of length '$L$' is rotated with an angular speed of '$\omega$' normal to a uniform magnetic field '$B$' about an axis passing through one end of the rod,as shown in the figure. The induced emf will be:
- A$\frac{1}{4} B^2 L \omega$
- B$\frac{1}{4} BL^2 \omega$
- C$\frac{1}{2} BL^2 \omega$
- D$\frac{1}{2} B^2 L^2 \omega$
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View SolutionIf current $I$ is flowing in a closed circuit with total resistance $R$,the rate of production of heat energy in the loop as we pull it with a constant speed $V$ is ($L=$ length of conductor,$B=$ magnetic field).
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