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Physics Question on Electromagnetic induction

A metal conductor of length 1 m1 \,m rotates vertically about one of its ends with an angular velocity 5 rad s−15 \,rad\, s^{-1}. If the horizontal component of earths magnetic field is 0.2×10−4 T0.2 \times10^{-4}\,T, then the emf developed between the ends of the conductor is

A

5 μ V5\, \mu\,V

B

5 m V5\,m\,V

C

50 μ V50\, \mu \, V

D

50 m V50\,m\,V

Answer

50 μ V50\, \mu \, V

Explanation

Solution

The emf developed between the ends of the conductor is
ε=12ωBl2=12×5×0.2×10−4×(1)2\varepsilon=\frac{1}{2}\omega Bl^{2}=\frac{1}{2}\times5\times0.2\times10^{-4}\times(1)^{2}
=5×10−5 V=50×10−6V=5\times10^{-5}\,V=50\times10^{-6}V
=50 μ V=50\, \mu\,V