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.
Path
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*}
Result
2. Determine the induced voltage in the coil during the change in current.
Path
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*}
Result
$ u_{\rm ind} = -197 ~\rm V$