A long straight solenoid with a cross-section | vedclass

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(A) According to Faraday's law of electromagnetic induction,the line integral of the electric field is equal to the negative rate of change of magneti

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$\frac{n \mu_0 a^2 \dot{I}}{2 r}$

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(A) According to Faraday's law of electromagnetic induction,the line integral of the electric field is equal to the negative rate of change of magnetic flux: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi}{dt}$.For a long solenoid,the magnetic field inside is $B = \mu_0 n I$. The magnetic flux through a circular loop of radius $r$ is $\phi = B \cdot A$.Case $1$: For $r Applying Faraday's law: $E(2 \pi r) = -\frac{d}{dt}(\mu_0 n I \pi r^2) = -\mu_0 n \pi r^2 \frac{dI}{dt}$.Thus,$E = -\frac{\mu_0 n r}{2} \frac{dI}{dt}$. The magnitude is $|E| = \frac{\mu_0 n r \dot{I}}{2}$.Case $2$: For $r > a$,the flux is only through the cross-section of the solenoid,so $\phi = (\mu_0 n I)(\pi a^2)$.Applying Faraday's law: $E(2 \pi r) = -\frac{d}{dt}(\mu_0 n I \pi a^2) = -\mu_0 n \pi a^2 \frac{dI}{dt}$.Thus,$E = -\frac{\mu_0 n a^2}{2 r} \frac{dI}{dt}$. The magnitude is $|E| = \frac{\mu_0 n a^2 \dot{I}}{2 r}$.

$A$ long straight solenoid with a cross-sectional radius $a$ and number of turns per unit length $n$ has a current varying with time as $I = I_0 \sin(\omega t)$ (or simply $dI/dt = I$). The magnitude of the electric field as a function of distance $r$ from the solenoid axis is,

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