Discussion
Consider an infinitely long solenoid with \(N\) turns per unit length, radius \(R\) and carrying a current \(I(t) = \alpha\cos\omega t\), where \(\alpha\) is a constant and \(\omega\) is the angular frequency. The magnitude of electric field at the surface of the solenoid is
Answer: A
The electric field at the surface is \(\frac{1}{2}\mu_{0}NR\omega\alpha\sin\omega t\) in magnitude.
Inside the solenoid \(B = \mu_{0}NI = \mu_{0}N\alpha\cos\omega t\). Faraday’s law around a circle of radius \(R\) gives
\[ \begin{aligned} E(2\pi R) &= -\frac{d}{dt}(\pi R^{2}B) \\ &= \pi R^{2}\mu_{0}N\alpha\omega\sin\omega t \\ E &= \tfrac{1}{2}\mu_{0}NR\omega\alpha\sin\omega t \end{aligned} \]
The options with \(\cos\omega t\) forget that the induced field depends on the rate of change of the current, which brings in \(\sin\omega t\); they also drop the amplitude \(\alpha\). \(\mu_{0}NR\omega\alpha\sin\omega t\) misses the factor \(\tfrac{1}{2}\) that comes from dividing the area \(\pi R^{2}\) by the circumference \(2\pi R\).
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