Solenoid with ramp current

Problem

A solenoid has $N = 800$ turns, length $\ell = 40 \ \mathrm{cm}$ and cross-section $S = 5 \ \mathrm{cm^2}$. The current grows linearly in time: $i(t) = 0{,}25 \ \mathrm{A/s} \cdot t$. Calculate the inductance of the solenoid and the self-induced EMF.

Solution

Inductance. For a solenoid $L = \dfrac{\mu_0 N^2 S}{\ell}$, with $\mu_0 = 4\pi \cdot 10^{-7} \ \mathrm{T \cdot m / A}$ and $S = 5 \cdot 10^{-4} \ \mathrm{m^2}$: $L = \frac{4\pi \cdot 10^{-7} \cdot (800)^2 \cdot 5 \cdot 10^{-4}}{0{,}40} = \ev{1{,}0 \cdot 10^{-3} \ \mathrm{H}}$

Self-induced EMF. The current grows at a constant rate $\dfrac{\dd i}{\dd t} = 0{,}25 \ \mathrm{A/s}$, so: $\mathcal{E} = -L \frac{\dd i}{\dd t} = -1{,}0 \cdot 10^{-3} \cdot 0{,}25 = \ev{-0{,}25 \ \mathrm{mV}}$ The minus sign indicates that the EMF opposes the growth of the current (Lenz’s law). Since $\dd i / \dd t$ is constant, the self-induced EMF is constant in time.

Links

Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione Objects: Solenoide