Question

# A metal conductor of length $1m$ rotates vertically about one of its ends at angular velocity $5$ radians per second. If the horizontal component of the earth's magnetic field is $0.2×{10}^{-4}T$, then the e.m.f developed between the two ends of the conductor is

A

$5\mu V$

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B

$50\mu V$

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C

$5mV$

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D

$50mV$

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Solution

## The correct option is B $50\mu V$Step 1. Given data:Length $\left(l\right)$ = $1m$ Angular velocity $\left(\omega \right)$= $5$ radians per second Horizontal component of the earth's magnetic field $\left(B\right)$ = $0.2×{10}^{-4}T$ Step 2. Formula used:Induced emf $\epsilon =B\omega {l}^{2}/2$Step 3. Calculations:Putting all given values, we get$\epsilon =\left(0.2×{10}^{-4}×5×1×1\right)/2\phantom{\rule{0ex}{0ex}}=50\mu V$Thus, the correct option is option B.

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