A helicopter has metallic blades with length 4 m extending outward from the central point and rotating at 3 rev/s. If the vertical component of Earth's magnetic field is 40 µT, then the emf induced between the blade tip and the central point is
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- 3.14 mV
- 2.83 mV
- 16 mV
- 6 mV
The Correct Option is D
Solution and Explanation
The emf induced in a rotating conductor in a magnetic field is given by \[\mathcal{E} = \frac{1}{2} B \omega l^2 = \frac{1}{2} B (2 \pi f) l^2 = \pi B f l^2,\]where $B$ is the magnetic field strength, $\omega$ is the angular velocity, $f$ is the frequency, and $l$ is the length of the conductor.
In this case, we are given that $l = 4 \, {m}$, $f = 3 \, {rev/s}$, and $B = 40 \, \mu {T} = 40 \times 10^{-6} \, {T}$, so \[\mathcal{E} = \pi B f l^2 = \pi \cdot 40 \times 10^{-6} \cdot 3 \cdot 4^2 \approx \boxed{6 \, {mV}}.\]
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