# 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) $50\; m V$

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)

$5\; \mu V$

Incorrect

Option 2)

$50\; \mu V$

Correct

Option 3)

$5\; m V$

Incorrect

Option 4)

$50\; m V$

Incorrect

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