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A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radian per second. If the horizontal component of earth’s magnetic field is 0.2 x 10-4 T, then the e.m.f. developed between the two ends of the conductor is

  • Option 1)

    5\; \mu V

  • Option 2)

    50\; \mu V

  • Option 3)

    5\; m V

  • Option 4)

    50\; m V


Answers (1)


As we learnt in 

Motional EMF -

\varepsilon = Blv

- wherein

B\rightarrow magnetic field

l\rightarrow length

v\rightarrow velocity of u perpendicular to uniform magnetic field.

Induced emf=\frac{1}{2}B\omega l^{2}=\frac{1}{2}\times (0.2\times 10^{-4})(5)(1)^{2}

\therefore \; \; Induced\; emf\, =\frac{10^{-4}}{2}=\frac{100\times 10^{-6}}{2}=50\mu V



Option 1)

5\; \mu V


Option 2)

50\; \mu V


Option 3)

5\; m V


Option 4)

50\; m V


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