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A metallic conductor of length $$1$$ m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity $$5$$ rad s$$^{-1}$$. If the horizontal component of earth's magnetic field is $$0.2 \times 10^{-4}$$ T, then emf induced between the two ends of the conductor is
Given:
$$Magnetic\ field\ B=0.2\times\ 10^{-4},\ length\ l=1m,\ angular\ velocity\ ω=5rad/s$$
Formula (emf in rotating rod):
$$E=\ \frac{\ 1}{2}Bωl^2$$
Calculation:
$$E=1/2\times0.2\times5\times(10^{-4})\times\ 1^2=0.5\times10−4V$$
$$E=50\times10^{-6}V=50μV$$
E=50×10−6 V=50 μV$$\mathcal{E} = 50 \times 10^{-6}\,$$
$$\text{V} = 50\,\mu\text{V}E=50×10−6V=50μV$$
Final Answer:
50 μV
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