Topic: Electromagnetics
A long solenoid of radius R, having N turns per unit length carries a time dependent current $I(t)=I_{0}\sin(\omega t)$. The magnitude of induced electric field at a distance $R/2$ radially from the axis of the solenoid is
- A. $\frac{R}{2}\mu_{0}NI_{0}\omega \cos(\omega t)$
- B. $\frac{R}{4}\mu_{0}NI_{0}\omega \cos(\omega t)$
- C. $\frac{R}{2}\mu_{0}NI_{0}\omega \sin(\omega t)$
- D. $\frac{R}{4}\mu_{0}NI_{0}\omega \sin(\omega t)$
Answer: B. $\frac{R}{4}\mu_{0}NI_{0}\omega \cos(\omega t)$
Explanation: According to Faraday's law, $\oint E \cdot dl = -\frac{d\phi}{dt}$. The magnetic flux $\phi$ enclosed by a loop of radius $P$ (where $P < R$) is $B \cdot A = (\mu_0 N I)(\pi P²)$. So, $E(2\pi P) = \mu_0 N \pi P² \frac{dI}{dt} \Rightarrow E = \frac{\mu_0 N P}{2} \frac{dI}{dt}$. Since $I(t) = I_0 \sin(\omega t)$, we have $\frac{dI}{dt} = I_0 \omega \cos(\omega t)$. At distance $P = R/2$, $E = \frac{\mu_0 N (R/2)}{2} I_0 \omega \cos(\omega t) = \frac{R}{4}\mu_0 N I_0 \omega \cos(\omega t)$.