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Question
What is the total flux linkage for a coil with \[N\] turns?
Options
\[N\Phi_B=NBA\cos(\omega t)\]
\[N\Phi_B=BA\cos(\omega t)\]
\[N\Phi_B=NBA\omega\sin(\omega t)\]
\[N\Phi_B=NBA\sin(\omega t)\]
MCQ
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Solution
Each of the \[N\] turns has flux \[BA\cos(\omega t)\]. Multiplying by the number of turns gives the total flux linkage.
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