Solution to 1996 Problem 46

We use the integral form of Faraday's Law,
\begin{align*}\int_P \mathbf{E} \cdot d \mathbf{l} = -\frac{d }{dt} \left(\int_S \mathbf{B} \cdot d \mathbf{a} \right) = \fra...
If we take the path P to be the circular wire loop (and we neglect the fact that it appears not to be a complete circle in the figure), then the left-hand side of the equation above is equal to the induced emf.
\begin{align*}\phi_B = \pi R^2 B \cos \theta \end{align*}
where \theta is the angle between the normal to the circular loop and the magnetic field. The wire loop rotates with an angular speed of \omega, therefore, \theta = \omega t + \delta where \delta is a constant.
\begin{align*}\left|\frac{d \phi_B}{dt} \right| = \left|\varepsilon\right| = \varepsilon_0 \sin \omega t\end{align*}
\begin{align*}\frac{d \phi_B}{dt} = - \omega \pi R^2 B \sin \left(\omega t + \delta \right)\end{align*}
\begin{align*}\left|\omega \pi R^2 B \sin \left(\omega t + \delta \right) \right| = \left| \varepsilon_0 \sin \omega t \right...
This equation can only hold if
\begin{align*}\left| \sin \left(\omega t + \delta \right) \right| = \left| \sin \omega t \right|\end{align*}
It follows then that
\begin{align*}\omega = \frac{\varepsilon_0}{\pi R^2 B}\end{align*}
Therefore, answer (C) is correct.

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