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— ee2:task_ljxf80q7vxywehqf_with_calculation [2024/07/15 22:23] (aktuell)
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 +{{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.
 +
 +#@HiddenBegin_HTML~ljxf80q7vxywehqf_11,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*}
 +
 +#@HiddenEnd_HTML~ljxf80q7vxywehqf_11,Path~@#
 +
 +#@HiddenBegin_HTML~ljxf80q7vxywehqf_12,Result~@#
 +$ L = 1.32 ~\rm mH$
 +#@HiddenEnd_HTML~ljxf80q7vxywehqf_12,Result~@#
 +
 +2. Determine the induced voltage in the coil during the change in current.
 +
 +#@HiddenBegin_HTML~ljxf80q7vxywehqf_21,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*}
 +#@HiddenEnd_HTML~ljxf80q7vxywehqf_21,Path~@#
 +
 +#@HiddenBegin_HTML~ljxf80q7vxywehqf_22,Result~@#
 +$ u_{\rm ind} = -197 ~\rm V$
 +#@HiddenEnd_HTML~ljxf80q7vxywehqf_22,Result~@#
 +
 +
 +#@TaskEnd_HTML@#