Physics equations/Faraday law/Faraday law example

Faraday's law of induction|Faraday's law of electromagnetic induction states that the induced electromotive force is the negative time rate of change of magnetic flux through a conducting loop.

${\displaystyle {\mathcal {E}}=-{{d\Phi _{B}} \over dt},}$

where ${\displaystyle {\mathcal {E}}}$ is the electromotive force (emf) in volts and Φ_B is the magnetic flux in Weber (Wb)|webers. For a loop of constant area, A, spinning at an angular velocity of ${\displaystyle \omega }$ in a uniform magnetic field, B, the magnetic flux is given by

${\displaystyle \Phi _{B}=B\cdot A\cdot \cos(\theta ),}$

where θ is the angle between the normal to the current loop and the magnetic field direction. Since the loop is spinning at a constant rate, ω, the angle is increasing linearly in time, θ=ωt, and the magnetic flux can be written as

${\displaystyle \Phi _{B}=B\cdot A\cdot \cos(\omega t).}$

Taking the negative derivative of the flux with respect to time yields the electromotive force.

${\displaystyle {\mathcal {E}}=-{\frac {d}{dt}}\left[B\cdot A\cdot \cos(\omega t)\right]}$	Electromotive force in terms of derivative
${\displaystyle =-B\cdot A{\frac {d}{dt}}\cos(\omega t)}$	Bring constants (A and B) outside of derivative
${\displaystyle =-B\cdot A\cdot (-\sin(\omega t)){\frac {d}{dt}}(\omega t)}$	Apply chain rule and differentiate outside function (cosine)
${\displaystyle =B\cdot A\cdot \sin(\omega t){\frac {d}{dt}}(\omega t)}$	Cancel out two negative signs
${\displaystyle =B\cdot A\cdot \sin(\omega t)\omega }$	Evaluate remaining derivative
${\displaystyle =\omega \cdot B\cdot A\sin(\omega t).}$	Simplify.