Self-induced EMF in a solenoid

Problem

In a solenoid the current varies as $I(t) = (5{,}0 \ \mathrm{A/s})\cdot t$. The self-inductance is $L = 0{,}50 \ \mathrm{H}$. Calculate the self-induced electromotive force.

Solution

The self-induced EMF is proportional to the rate of change of the current: $\mathcal{E} = -L\,\ddt{I} = -0{,}50 \cdot 5{,}0 = \ev{-2{,}5 \ \mathrm{V}}$ The negative sign (Lenz’s law) indicates that the EMF opposes the increase in current.

Collegamenti

Argomenti: Electromagnetic induction Concetti: Inductance and self-induction Oggetti: Solenoid