Exercise E8 Self-Induction
(written test, approx. 8 % of a 120-minute written test, SS2024)
A coil with a length of $30 ~\rm cm$ and a radius of $2 ~\rm cm$ has $500$ turns.
The current through the coil changes linearly from $0$ to $3 ~\rm A$ in $0.02 ~\rm ms$.
The arrangement is located in air ($\mu_{\rm r}=1$).
$\mu_0= 4\pi \cdot 10^{-7} ~\rm Vs/Am$
1. Calculate the (self-)inductance of the coil.
The formula for the induction of a long coil is:
\begin{align*}
L &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A}\over{l}} \\
&= 4\pi \cdot 10^{-7} {~\rm Vs/Am} \cdot (500)^2 \cdot {{\pi \cdot (2\cdot 10^{-2} ~\rm m)^2}\over{ 2 \cdot 10^{-2} ~\rm m}} \\
\end{align*}
$ L = 1.32 ~\rm mH$
2. Determine the induced voltage in the coil during the change in current.
For the linear change of the current the formula of the induced voltage can also be linearized:
\begin{align*}
u_{\rm ind} &= - L \cdot {{ {\rm d} i }\over{ {\rm d} t }} \\
&\rightarrow - L \cdot {{ {\Delta} i }\over{ {\Delta} t }} \\
&= - 1.32 \cdot 10^{-3} \cdot {{3 A}\over{0.02 \cdot 10^{-3} s}}
\end{align*}
$ u_{\rm ind} = -197 ~\rm V$