{{tag>self:induction exam_ee2_SS2024}}{{include_n>1070}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Self-Induction \\
<fs medium>(written test, approx. 8 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#

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.

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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*}

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$ L = 1.32 ~\rm mH$
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2. Determine the induced voltage in the coil during the change in current.

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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*}
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$ u_{\rm ind} = -197 ~\rm V$
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