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https://texercises.raemilab.ch/exercise/rotor-blades/
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Exercise:
The RO long rotor blades of a helicopter turn at fO. The vertical component of the earth’s magnetic field at the position of the helicopter has a magnitude of BO. Calculate the induced emf between the rotor axis and the tip of one blade.

Solution:
In order to use Faraday's law we have to choose an arbitrary reference position e.g. the dashed line in the figure below. The magnetic flux through the shaded area is given by Phi_m B At B R^pifracomega tpi BfracR^ omega t R is the length of a rotor blade. center includegraphicswidthmm#image_path:rotor-blade# center It follows for the induced emf mathcalE -dotPhi_m -fracB R^omega VF -Btimes R^timespitimesf V approx resultVP- The voltage can also be derived using the expression for the motional emf but we have to take o account that the different parts of the rotor blade move at different speeds: vr omega r The induced emf for a segment with length ddr at a distance r from the rotational axis is ddmathcalE B vr ddr B omega r ddr The total emf can then be calculated by egrating from r_ to r_R: mathcalE _^RB omega r mathopdr B omega _^Rr mathopdr B omegafracleftr^right_^R fracBomega R^ This is obviously the same result but for the sign found using Faraday's law.
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Exercise:
The RO long rotor blades of a helicopter turn at fO. The vertical component of the earth’s magnetic field at the position of the helicopter has a magnitude of BO. Calculate the induced emf between the rotor axis and the tip of one blade.

Solution:
In order to use Faraday's law we have to choose an arbitrary reference position e.g. the dashed line in the figure below. The magnetic flux through the shaded area is given by Phi_m B At B R^pifracomega tpi BfracR^ omega t R is the length of a rotor blade. center includegraphicswidthmm#image_path:rotor-blade# center It follows for the induced emf mathcalE -dotPhi_m -fracB R^omega VF -Btimes R^timespitimesf V approx resultVP- The voltage can also be derived using the expression for the motional emf but we have to take o account that the different parts of the rotor blade move at different speeds: vr omega r The induced emf for a segment with length ddr at a distance r from the rotational axis is ddmathcalE B vr ddr B omega r ddr The total emf can then be calculated by egrating from r_ to r_R: mathcalE _^RB omega r mathopdr B omega _^Rr mathopdr B omegafracleftr^right_^R fracBomega R^ This is obviously the same result but for the sign found using Faraday's law.
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Attributes & Decorations
Branches
Magnetism
Tags
electromagnetic induction, faraday law, induced emf, magnetic flux, motional emf
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Difficulty
(2, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration